problem solving triangle

• What is Problem-Oriented Policing?
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• Key Elements of POP

The SARA Model

• The Problem Analysis Triangle
• Situational Crime Prevention
• 25 Techniques
• Links to Other POP Friendly Sites
• About POP en Español

A commonly used problem-solving method is the SARA model (Scanning, Analysis, Response and Assessment). The SARA model contains the following elements:

• Identifying recurring problems of concern to the public and the police.
• Identifying the consequences of the problem for the community and the police.
• Prioritizing those problems.
• Developing broad goals.
• Confirming that the problems exist.
• Determining how frequently the problem occurs and how long it has been taking place.
• Selecting problems for closer examination.
• Identifying and understanding the events and conditions that precede and accompany the problem.
• Identifying relevant data to be collected.
• Researching what is known about the problem type.
• Taking inventory of how the problem is currently addressed and the strengths and limitations of the current response.
• Narrowing the scope of the problem as specifically as possible.
• Identifying a variety of resources that may be of assistance in developing a deeper understanding of the problem.
• Developing a working hypothesis about why the problem is occurring.
• Brainstorming for new interventions.
• Searching for what other communities with similar problems have done.
• Choosing among the alternative interventions.
• Outlining a response plan and identifying responsible parties.
• Stating the specific objectives for the response plan.
• Carrying out the planned activities.

Assessment:

• Determining whether the plan was implemented (a process evaluation).
• Collecting pre and postresponse qualitative and quantitative data.
• Determining whether broad goals and specific objectives were attained.
• Identifying any new strategies needed to augment the original plan.
• Conducting ongoing assessment to ensure continued effectiveness.

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Triangles – example 1:.

Use the area formula: Area \(= \color{blue}{\frac{1}{2 }(base \ × \ height)}\) base \(=10\) and height \(=6\) Area \(= \frac{1}{2} (10×6)=\frac{1}{2} (60)=30\)

Triangles – Example 2:

What is the missing angle of the following triangle?

All angles in a triangle sum up to \(180\) degrees. Then: : \(45+60+x=180 → 105+x=180 → x=180-105=75\)

Triangles – Example 3:

What is the area of the following triangle?

Use the area formula: Area \(= \color{blue}{\frac{1}{2 }(base \ × \ height)}\) base \(=12\) and height \(=8\) Area \(= \frac{1}{2} (12×8)= \frac{1}{2} (96)=48 \)

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Triangles – example 4:.

What is the area of a triangle with base \(6\) and height \(5\)?

Use the area formula: Area \(= \color{blue}{\frac{1}{2 }(base \ × \ height)}\) base \(=6\) and height \(=5\) Area \(= \frac{1}{2 } (5×6)= \frac{1}{2} (30)=15 \)

Exercises for Solving Triangles

Find the measure of the unknown angle in each triangle., download triangles worksheet.

• \(\color{blue}{45^{\circ}}\)
• \( \color{blue}{ 15^{\circ}}\)
• \( \color{blue}{ 40^{\circ}}\)

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Center for Problem-Oriented Policing

• » What is Problem-Oriented Policing?
• » History of Problem-Oriented Policing
• » Key Elements of POP
• » The SARA Model
• « The Problem Analysis Triangle
• » Situational Crime Prevention
• » 25 Techniques
• » Links to Other POP Friendly Sites
• » About POP en Español

The Problem Analysis Triangle

While the SARA model is useful as a way of organizing the approach to recurring problems, it is often very difficult to figure out just exactly what the real problem is. The problem analysis triangle (sometimes referred to as the crime triangle) provides a way of thinking about recurring problems of crime and disorder. This idea assumes that crime or disorder results when (1) likely offenders and (2) suitable targets come together in (3) time and space, in the absence of capable guardians for that target. A simple version of a problem analysis triangle looks like this:

Offenders can sometimes be controlled by other people: those people are known as handlers. Targets and victims can sometimes be protected by other people as well: those people are known as guardians. And places are usually controlled by someone: those people are known as managers. Thus, effective problem-solving requires understanding how offenders and their targets/victims come together in places, and understanding how those offenders, targets/victims, and places are or are not effectively controlled. Understanding the weaknesses in the problem analysis triangle in the context of a particular problem will point the way to new interventions. A complete problem analysis triangle looks like this:

Problems can be understood and described in a variety of ways. No one way is definitive. They should be described in whichever way is most likely to lead to an improved understanding of the problem and effective interventions. Generally, incidents that the police handle cluster in four ways:

• Behavior . Certain behavior(s) is (are) common to the incidents. For example, making excessive noise, robbing people or businesses, driving under the influence, crashing vehicles, dealing drugs, stealing cars. There are many different behaviors that might constitute problems.
• Place . Certain places can be common to incidents. Incidents involving one or more problem behaviors may occur at, for example, a street corner, a house, a business, a park, a neighborhood, or a school. Some incidents occur in abstract places such as cyberspace, on the telephone, or through other information networks.
• Persons . Certain individuals or groups of people can be common to incidents. These people could be either offenders or victims. Incidents involving one or more behaviors, occurring in one or more places may be attributed to, for example, a youth gang, a lone person, a group of prostitutes, a group of chronic inebriates, or a property owner. Or incidents may be causing harm to, for example, residents of a neighborhood, senior citizens, young children, or a lone individual.
• Time . Certain times can be common to incidents. Incidents involving one or more behaviors, in one or more places, caused by or affecting one or more people may happen at, for example, traffic rush hour, bar closing time, the holiday shopping season, or during an annual festival.

There is growing evidence that, in fact, crime and disorder does cluster in these ways. It is not evenly distributed across time, place, or people. Increasingly, police and researchers are recognizing some of these clusters as:

• Repeat offenders attacking different targets at different places.
• Repeat victims repeatedly attacked by different offenders at different places.
• Repeat places (or hot spots) involving different offenders and different targets interacting at the same place.

The Problem Analysis Triangle was derived from the routine activity approach to explaining how and why crime occurs. This theory argues that when a crime occurs, three things happen at the same time and in the same space:

• a suitable target is available.
• there is the lack of a suitable guardian to prevent the crime from happening.
• a motivated offender is present.

Check out the list of readings under the POP Center recommended readings .

El Triangulo de Análisis de Problemas (Triángulo del Delito)

El Triángulo de Análisis de Problemas o más comúnmente conocido como el Triángulo del Delito, proporciona una forma de evaluar y analizar los problemas recurrentes del delito y de la alteración del orden público. Esta idea se fundamenta en que el delito o la alteración del orden público es una consecuencia producida por: (1) La acción o accionar de delincuentes (2) se conduce sobre objetivos adecuados, convenientes o lucrativos (3) se ejecuta en un tiempo y en un lugar determinado, en la ausencia de un protector/guardián competente del objetivo. Una versión sencilla de un triángulo de análisis de problemas se observa de la siguiente forma:

Los delincuentes pueden estar controlados a veces por otras personas; a estas personas se les denomina en el presente modelo como manipuladores. Los objetivos y las víctimas pueden estar protegidos a veces también por otras personas; a estas personas se las define como protectores/guardianes. Los lugares normalmente pueden estar controlados por alguien; a estas personas se las define como administradores/gerentes. De modo que, el desarrollo efectivo de solución de problemas (en nuestro caso problemas delictivos) pasa por entender como los delincuentes y sus víctimas/objetivos convergen en un lugar; también entender como esa convergencia de los delincuentes, las víctimas/objetivos, en unos lugares no está efectivamente controlada. Entendiendo las vulnerabilidades que se observan en el triángulo de análisis de problemas en el contexto de un problema en particular, nos indicará el camino a seguir para desarrollar nuevas formas de intervención. Una versión del triángulo de análisis de problemas completo se presenta y luce de la forma siguiente:

Los problemas delictivos pueden entenderse y describirse en diversas formas. Es importante destacar que ninguna forma es definitiva. Los problemas debieran describirse de forma tal, que facilite su comprensión y que a su vez esta comprensión origine formas efectivas de intervención del problema. Generalmente, los incidentes que la policía maneja se pueden agrupar de estas cuatro maneras:

• Conducta: Algunas conductas son comunes a los incidentes, por ejemplo: Robar a personas comunes o a comerciantes, manejar bajo la influencia de drogas, chocar vehículos, vender drogas ilícitas, robar vehículos. Hay diversas formas de conductas que pueden constituir un problema.
• Lugar: Algunos lugares pueden ser comunes a los incidentes. Incidentes donde se involucran uno o más problemas de conducta pueden ocurrir por ejemplo en: Una esquina, en una casa, en un comercio o negocio, en un parque, en un vecindario o en una escuela. Algunos incidentes pueden ocurrir en lugares abstractos como el ciberespacio, en el teléfono, o a través de otras redes de información.
• Personas: Ciertas personas o grupos de personas pueden ser comunes a los incidentes. Estas personas pueden ser tanto víctimas como delincuentes. Incidentes donde se involucren una o más conductas, que ocurren en uno o más lugares pueden atribuirse a, por ejemplo: Una pandilla juvenil, una persona solitaria, un grupo de prostitutas, un grupo de alcohólicos, o el propietario de una casa. O también, incidentes donde se cause un daño por ejemplo a: Los residentes de un vecindario, a personas ancianas, niños, o a individuos solitarios.
• Tiempo: Algunos tiempos pueden ser comunes a los incidentes. Incidentes donde se involucren una o mas conductas, en uno o mas lugares, causado por o afectando a una o mas personas pueden ocurrir por ejemplo en: La hora pico del tráfico automotor, a la hora de cerrar bares o discotecas, al comienzo de las fechas de las vacaciones Decembrinas/Carnavales/Semana Santa o durante un día de fin de año.

Si bien es cierto que existe una creciente evidencia que indica de hecho, que el delito y las alteraciones del orden público se agrupan en las cuatro áreas descritas previamente, no hay evidencias de que exista una distribución igualitaria entre el tiempo, el lugar o las personas entre los incidentes; lo que si han reconocido de forma creciente las policías y los investigadores, es que las características de los incidentes se pueden agrupar de la siguiente forma:

• Repetidamente los mismos delincuentes atacan diferentes objetivos en diferentes lugares.
• Repetidamente las mismas víctimas son atacadas por diferentes delincuentes en diferentes lugares.
• Repetidamente en los mismos lugares (zonas rojas) interactúan en el mismo lugar diferentes delincuentes en contra de diferentes objetivos.

El Triángulo de Análisis de Problemas fue desarrollado como una actividad rutinaria para enfocar y explicar como y cuando ocurre el delito. La presente teoría argumenta que cuando un delito ocurre, tres cosas convergen a la misma hora y en el mismo lugar:

• Existe un objetivo conveniente/apropiado/lucrativo está disponible.
• No existe un guardián/protector que efectivamente prevenga la ocurrencia del delito.
• Existe un delincuente que está motivado y presente.
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Trig without Tears Part 4:

Solving Triangles

Copyright © 1997–2024 by Stan Brown, BrownMath.com

For all of this, you need only two tools, the Law of Sines and the Law of Cosines . The Law of Sines relates any two sides and the angles opposite them, and the Law of Cosines relates all three sides and one angle.

• How to Solve Triangles on TI-83/84 includes a TI-83/84 program to automate the computations mentioned in this chapter.
• There’s an online triangle solver , by Jesús SD, for checking your answers. Click through the prompt about Chrome; it seems to work fine in the other browsers that I tried.

Law of Sines

Law of cosines, special note: side-side-angle, solving triangles from area, practice problems, btw: great book on problem solving.

Let’s look at a specific example to start with. Suppose you have a triangle where one side has a length of 180, an adjacent angle is 42°, and the opposite angle is 31°. You’re asked to find the other angle and the other two sides.

We’ve already got the diagram, but let’s see if those other techniques will be helpful. (By the way, they’re not original with me, but are from a terrific book on problem-solving techniques that I think you should know about.)

“Can you use what you already know to solve a piece of this problem?” For example, if this were a right triangle you’d know right away how to write down the lengths of sides in terms of sines or cosines .

But it’s not a right triangle, alas. Is there any way to turn it into a right triangle? Not exactly, but if you construct a line at right angles to one side and passing through the opposite vertex, you’ll have two right triangles. Maybe solving those right triangles will show how to solve the original triangle.

Let’s start by writing down things we know that relate the sides and angles of the two right triangles in the diagram above. You remember how to write down the lengths of the legs of a right triangle ? The leg is always equal to the hypotenuse times either the cosine of the adjacent angle or the sine of the opposite angle. (If that looks like just empty words to you, or even if you’re not 100% confident about it, please go back and review that section until you feel confident.)

AD =  b  cos  A and CD =  b  sin  A

By the same reasoning, in the other triangle you have

DB =  a  cos  B and CD =  a  sin  B

This is striking: you see two different expressions for the length CD. But things that are equal to the same thing are equal to each other. That means that

b  sin  A  = a  sin  B

Divide through by sin  A and you have the solution for the general case:

b  =  a  sin  B  / sin  A

b  = 180 × sin 42° / sin 31° ≈ 234

What about the third angle, C , and the third side, c ? Well, when you have two angles of a triangle you can find the third one easily:

A  + B  + C  = 180°

C  = 180° − A  − B

In this case, C  = 180° − 31° − 42° = 107°.

For the third side, there are a couple of ways to go. You wrote expressions above for AD and DB, and you know that c  = AD+DB, so you could compute c  =  b  cos  A  + a  cos  B .

But that’s two multiplies and an add, a bit more complicated than the one multiply and one divide to find side b . I’m lazy, and I like to reduce the amount of tapping I do on my calculator. Is there an easier way, even if just slightly easier? Yes, there is. Go back a step, to

a  sin  B = b  sin  A

Divide left and right by (sin  A )(sin  B ) to get

a /sin  A  =  b /sin  B

Because C  > 90°, this perpendicular happens to be outside the triangle and the two right triangles ABD and CBD overlap. But this won’t affect the algebra. By the way, the angle in triangle CBD is not C but 180° − C , the supplement of C . Angle C belongs to the original triangle ABC.

You can write the length of the common side BD as

BD = c  sin  A (in triangle ABD)

BD = a  sin(180° − C ) (in triangle CBD)

But sin(180° − C ) = sin  C , so you have

BD = a  sin  C (in triangle CBD)

Set the two computed lengths of BD equal to each other, and divide by (sin  A )(sin  C ):

a  sin  C  = c  sin  A

a /sin  A  = c /sin  C

But we already figured out earlier that

Combining these two equations you have the Law of Sines :

(28) Law of Sines—First Form:

a /sin  A  =  b /sin  B  =  c /sin  C

This is very simple and beautiful: for any triangle, if you divide any side by the sine of the opposite angle , you’ll get the same result. This law is valid for any triangle.

You can derive the Law of Sines at need, so I don’t specifically recommend memorizing it. But it’s so simple and beautiful that it’s pretty hard not to memorize if you use it at all. It’s also pretty hard to remember it wrong: there are no alternating plus and minus signs or combinations of different functions.

a /sin  A = c /sin  C

a (sin  C )/(sin  A ) = c

c = 180 × (sin 107°)/(sin 31°) ≈ 334

The Law of Sines is sometimes given upside down:

(29) Law of Sines—Second Form:

(sin  A )/ a  =  (sin  B )/ b  =  (sin  C )/ c

Of course that’s the same law, just as 2/3 = 6/9 and 3/2 = 9/6 are the same statement. Work with it either way, and you’ll come up with the same answers.

In most cases where you use the Law of Sines, you get a unique solution. But sometimes you get two solutions (or none) in the side-side-angle case, where you know two sides and an angle that’s not between them. Please see the Special Note below, after the table.

The Law of Sines is fine when you can relate sides and angles. But suppose you know three sides of the triangle—for instance a  = 180, b  = 238, c  = 340—and you have to find the three angles. The Law of Sines is no good for that, because it relates two sides and their opposite angles. If you don’t know any angles, you have an equation with two unknowns and you can’t solve it.

Let’s look back at that generic triangle with a perpendicular dropped from vertex C . You may remember that when we first looked at this picture, we pulled out information using both the sine and the cosine of the two angles. We used the sine information to develop the Law of Sines, but we never went anywhere with the cosine information, which was

AD = b cos  A    and    DB or BD = a cos  B

Let’s see where that can lead us. You remember that the way we came up with the Law of Sines was to write two equations that featured the length of the construction line CD, and then combine the equations to eliminate CD. Can we do anything like that here?

Well, we know the other two sides of those right triangles, so we can write an expression for the height CD using the Pythagorean theorem—actually, two expressions, one for each triangle.

a ² = (CD)² + (BD)² ⇒ (CD)² =  a ² − (BD)²

b ² = (CD)² + (AD)² ⇒ (CD)² =  b ² − (AD)²

and therefore

a ² − (BD)² = b ² − (AD)²

Substitute the known values BD = a  cos  B and AD = b  cos  A , and you have

a ² − a ² cos²  B  =  b ² − b ² cos²  A

Bzzt! No good! That uses two sides and two angles, but we need an equation in three sides and one angle, so that we can solve for that angle. Let’s back up a step, to a ² − (BD)² =  b ² − (AD)², and see if we can go in a different direction.

Maybe the problem is in treating BD and AD as separate entities when actually they’re parts of the same line. Since BD + AD =  c , we can write

BD = c  − AD

BD = c  − b  cos  A .

Notice that this brings in the third side, c , and angle B drops out. Substituting, we now have

a ² − ( c − b  cos  A )² = b ² − ( b  cos  A )²

This looks worse than the other one, but actually it’s better because it’s what we’re looking for: an equation for the three sides and one angle. We can solve it with a little algebra:

a ² − c ² + 2 b c cos  A − b ²cos² A = b ² − b ²cos² A

a ² − c ² + 2 b c cos  A = b ²

2 b c cos  A = b ² + c ² − a ²

cos  A = ( b ² + c ² − a ²) / 2 b c

cos  A = (238² + 340² − 180²) / (2 × 238 × 340)

cos  A ≈ 0.864088

A ≈ 30.2°

Do the same thing to find the second angle (or use the Law of Sines, since it’s less work), then subtract the two known angles from 180° to find the third angle.

You can find the Law of Cosines for the other angles by following the same process using the other two perpendiculars.

(30) Law of Cosines—First Form:

cos  B = ( a ² + c ² − b ²) / 2 a c

cos  C = ( a ² + b ² − c ²) / 2 a b

cos  C = (180² + 238² − 340²) / (2 × 180 × 238)

cos  C ≈ −0.309944

C ≈ 108.1°

Notice that the Law of Cosines automatically handles acute and obtuse angles. Remember from the diagram in Functions of Any Angle that cos  A is negative when A is between 90° and 180°. Because the cosine has unique values all the way from 0° to 180°, you never have to worry about multiple solutions of a triangle when you use the Law of Cosines.

There’s another well-known form of the Law of Cosines, which may be a bit easier to remember. Start with the above form, multiply through by 2 a b , and isolate c on one side:

2 a b  cos  C = a ² + b ² − c ²

c ² = a ² + b ² − 2 a b  cos  C

You can play the same game to solve for the other two sides:

(31) Law of Cosines—Second Form:

a ² = b ² + c ² − 2 b c  × cos  A

b ² = a ² + c ² − 2 a c  × cos  B

c ² = a ² + b ² − 2 a b  × cos  C

Typically you’ll use the Law of Cosines in the first form for finding an angle and the second form for finding a side .

Probably you don’t want to try to remember that, but it’s not as hard as it looks. I think of it this way: the square of one side is the sum of the squares of the other two, like Pythagoras, but with a “correction factor” of 2 times those same sides times the cosine of the opposite angle.

Detective Work: Solving All Types of Triangles

With just the definitions of sine, cosine, and tangent , you can solve any right triangle . If you’ve got the Law of Sines and the Law of Cosines under your belt, you can solve any triangle that exists . (Some sets of givens lead to an impossible situation, like a “triangle” with sides 3-4-9.)

Really, it’s pretty straightforward. Whenever you have to solve a triangle, think about what you have and then think about which formula you can use to get what you need. (When you have two angles, you can always find the third by A  + B  + C  = 180°.)

Many people find it easier to think about the known elements of a triangle as a “case”. For instance, if you know two angles and the side between them, that’s case ASA; if you know two angles and a side that’s not between them, that’s case AAS, and so on.

I’m not presenting the following table for you to memorize. Instead, what I hope to do is show you that between the Law of Sines and the Law of Cosines you can solve any triangle, and that you simply pick which law to use based on which one has just one unknown and otherwise uses information you already have.

Most cases can be solved with the Law of Sines. But if you have three sides ( SSS ), or two sides and the angle between them ( SAS ), you must begin with the Law of Cosines.

Suppose you know acute angle B and sides a and b . Given those facts, there are two different ways you could draw the triangle, as shown in the picture. How can this be? Well, you use the Law of Sines to find angle A . Let’s say you find sin  A  = 0.95. That means A could be either about 72° or the supplement, about 108° . Remember that the sine of any angle and the sine of its supplement are the same.

This is the infamous ambiguous case . You can see the problem from the picture: the known opposite side b can take either of two positions that satisfy the given the lengths of a and b . Those two positions give rise to two different values for angle A , two different values for angle C , and two different values for side c . Think about it for a while, maybe sketch some pictures, and you’ll see that this ambiguity can arise only when the known angle is acute, and the adjacent side is longer than the opposite side, and the opposite side is greater than the height.

(Once you know the size or sizes of angle A , you can find angle C by subtracting 180° − B  − A .)

Here’s a complete rundown of all the possibilities with the SSA case:

For heaven’s sake, don’t try to memorize that table! Instead, always draw a picture. If you can draw two pictures that both fit all the available facts, you have two legitimate solutions. If only one picture fits all the facts, it will show you which angle (if any) is > 90°. And if you can’t make any picture that fits the facts, the triangle has no solution.

If you do have two solutions, what do you do? If you have no other information to go on, of course you report both solutions. But check the situation carefully. Maybe you’re told explicitly which is the largest angle, or it’s implied by other facts you know. In that case your solution is constrained, and you reject the solution that doesn’t meet the constraints.

Example: Suppose you are asked to solve a triangle with B  = 36.9° a  = 75.3, and b  = 51.3. How do you proceed?

This is the side-side-angle case: you know two sides a and b , and a non-included angle B . The adjacent side to angle B , a  = 75.3, is larger than the opposite side, b  = 51.3, so you have to compute the height, h = 75.3 sin 36.9° ≈ 45.2. The opposite side, b  = 51.3, is larger than this, so there are two solutions.

Use the Law of Sines, equation 29 , to get the second angle:

(sin  A )/ a = (sin  B )/ b

sin  A = ( a / b ) sin  B

sin  A = (75.3 / 51.3) sin 36.9° ≈ 0.8813

A = 61.8° or 180° − 61.8° = 118.2°

Given: Area and Two Angles

In April 2016, Caroline McKnoe asked me how to solve a triangle if you have two angles and the area . I hadn’t run across that one before, but it’s doable with the standard ploy of dropping a perpendicular.

area = ( c b sin  A )/2

We know angle A —even if A isn’t one of the two givens we can easily find it by subtracting the other two from 180°—but there are two unknown sides in that equation. How can we eliminate one of them? We need some second equation that involves b and c but no other unknowns. The answer is in the Law of Sines :

b /sin  B  = c /sin  C ⇒ b  = c  sin  B /sin  C

Substitute that in the equation for area:

area = ( c ² sin  B  sin  A )/(2 sin  C )

Solve for side c :

c ² = 2 area sin  C /(sin  A  sin  B )

(32) c = √ 2 × area × sin  C /(sin  A  sin  B )

Finally, use the Law of Sines to find sides a and b .

Given: Area and Two Sides

In the previous section, we found a formula for area in terms of two sides and the included angle:

We couldn’t use that directly when we knew two angles and the area, but if we know two sides and the area then this formula is exactly what we want. Just solve for sin  A :

(33) sin  A = 2 × area/( b c )

Next, use the Law of Cosines to find side a . Finally, use the Law of Sines or Law of Cosines to find a second angle, and subtract those angles from 180° to find the third angle.

To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.

Recommendation : Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.

You’ll find full solutions for all problems . Don’t just check your answers, but check your method too.

How long must the monorail and the foot bridge be?

Bonus question: If the river has the same width all along the stretch from A to B , how wide is it?

You’re now driving along a straight side road. At the end of 9.8 miles on the side road, you turn 135° to the right, on a third road. (If you’re visualizing this from above, the 135° change of direction corresponds to an angle of 180° − 135° = 45° in the triangle.)

Assuming that road continues in the same direction, how far must you drive to reach your starting point?

I have to recommend a terrific little book, How To Solve It by G. Polya. Most teachers aren’t very good at teaching you how to solve problems and do proofs. They show you how they do them, and expect you to pick up their techniques by a sort of osmosis. But most of them aren’t very good at explaining the thought process that goes into doing a geometrical proof, or solving a dreaded “story problem”.

Polya’s book does a great job of teaching you how to solve problems. He shows you the kinds of questions you should ask yourself when you see a problem. In other words, he teaches you how to get yourself over the hum, past the floundering that most people do when they see an unfamiliar problem. And he does it with lots of examples, so that you can develop confidence in your techniques and compare your methods with his. The techniques I’ve mentioned above are just three out of the many in his book.

There’s even a handy checklist of questions you can ask yourself whenever you’re stuck on a problem.

How To Solve It was first published in 1945, and it’s periodically in and out of print. If you can’t get it from your bookstore, go to the library and borrow a copy. You won’t be sorry.

My thanks also to Alex for pointing out in the SSA example where I twice wrote side a where I meant b .

• 19 Nov 2021 : Updated a link here .
• 19 Feb 1997 : New document.

next:  5/Functions of Any Angle

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• Solving ...

Solving Triangles

In this tutorial, we’ll learn all about solving triangles. So let’s begin by looking at what it actually means to solve a triangle.

There are six values describing the six parts of a triangle — three sides and three internal angles.

Now, if we know any three of those six values, we can find the other three (with one exception that we’ll look at later). And that’s what solving triangles is all about.

To solve a triangle is to find its unknown (or required) side lengths and angles using what is known about the triangle.

Solving Triangles — Important Concepts

Before we dive into the different types of problems and examples and start solving them, here are the important concepts, formulas, and relations that are essential to being able to solve such problems.

So first things first. When naming sides and angles, a side is represented by the small case version of the letter for the opposite angle.

For example, the angle opposite to the side A B \hspace{0.2em} AB \hspace{0.2em} A B is angle C \hspace{0.2em} C \hspace{0.2em} C . So it (or its length) is represented by c \hspace{0.2em} c \hspace{0.2em} c .

Sum of Angles

The sum of the three internal angles of a triangle is 180 ° \hspace{0.2em} 180 \degree 180° .

Law of Cosines

The law of cosines gives the relationship between the side lengths of a triangle and the cosine of any of its angles. It says —

Re-framing the formula for other sides, we have

For cases where we need to find angles using the cosine rule, the three formulas can be rearranged as —

Law of Sines

The law of sines states that the ratio of side length to the sine of opposite angle is the same for all sides in a triangle.

When we find an angle using its sine value, there are two possibilities. The angle can be acute or obtuse (angle greater than 90 ° \hspace{0.2em} 90 \degree 90° ).

For example,

So which angle do we choose? Here are a couple of points to keep in mind.

A triangle can have a maximum of one obtuse angle. Otherwise, the sum would exceed 180 ° \hspace{0.2em} 180 \degree 180° .

The obtuse angle (if present) would be the angle opposite the largest side (itself being the largest angle).

If it sounds confusing, don’t worry. Wait till we get to the examples.

The Different Types

Depending on which three of the six values are known to us, the problems under the topic of solving triangles can be divided into five categories.

1. Side-Side-Side (SSS)

In this type, all three sides are known to us.

To solve, start by finding one angle using the cosine rule. Then find the two remaining angles using the sine rule.

2. Side-Angle-Side (SAS)

Here, two sides and the included angle (the angle contained between them) are known to us.

To solve, first find the third side using the law of cosines. Then find the two unknown angles using the law of sines.

3. Side-Side-Angle (SSA)

In this case, two sides and one angle (not included) are known.

(angle sum rule). Finally, find the missing side using the sine rule.

4. Angle-Side-Angle (ASA)

Here, two angles and the included side (the side shared by the two angles) are known to us.

To solve, start by finding the third side using the angle sum property. Then find the two unknown sides using the sine rule.

5. Angle-Angle-Side (AAS)

In this type, two angles and one side (not included) are known.

The steps to solve this are the same as those for the ASA type. The first step is to find the third angle using the angle sum property. Then you find the two unknown sides using the sine rule.

6. Angle-Angle-Angle (AAA)

In this case, the three angles are known to us.

You can’t solve this type. Here's why.

A triangle is not uniquely determined by its three angles. For the same three angles, you can have infinitely many triangles. They will have the same shape but different sizes ( similar triangles ).

For example, here are two triangles with the same internal angles (so similar by AA criteria) but with different size.

So how would one know which triangle it is without knowing at least one side-length?

Solving Triangles — Examples

Alright, now it’s time to use what we have learned so far and solve examples covering the different types discussed above.

The lengths of the three sides of a triangle are 6 \hspace{0.2em} 6 \hspace{0.2em} 6 , 3 \hspace{0.2em} 3 \hspace{0.2em} 3 , and 4 \hspace{0.2em} 4 \hspace{0.2em} 4 . Find the measures of its angles.

Step 0.  We start by drawing a rough sketch of the triangle and labeling the information given in the question. It’s not necessary but often makes things easier and helps avoid silly mistakes.

Step 1.  Use the Cosine Rule to find the largest angle

When we know all the side lengths, we can use the Cosine Rule to find any of the angles.

However, I strongly recommend you find the largest angle first — the angle opposite to the longest side. Why?

Because if there is an obtuse angle ( > 90 ° ) \hspace{0.2em} (>90 \degree) \hspace{0.2em} ( > 90° ) in the triangle, it has to be this angle. So once this is out of the way, we can use the sine rule to find the other angles without having to worry about the obtuse options.

Here the largest angle would be A \hspace{0.2em} A \hspace{0.2em} A . So using the formula for cos ⁡ A \hspace{0.2em} \cos A \hspace{0.2em} cos A , we have

Taking cos inverse on both sides.

Step 2.  Use the Sine Rule for one of the remaining angles

Now that we know the three sides and one angle, we can use the Sine Rule to find any of the remaining two angles. Let's go for B \hspace{0.2em} B \hspace{0.2em} B .

Taking sine inverse on both sides

Step 3.  Use the Angle Sum Property to find the third angle

And we have solved the triangle.

Note — We could have found the third angle ( C \hspace{0.2em} C \hspace{0.2em} C ) using the sine rule, as we did for angle B \hspace{0.2em} B \hspace{0.2em} B . I used the angles sum property just because the calculation was easier. You can also use one to find the angle and the other to validate your answer(s).

Two sides and the included angle in a triangle measure 12 \hspace{0.2em} 12 \hspace{0.2em} 12 , 14 \hspace{0.2em} 14 \hspace{0.2em} 14 , and 110 ° \hspace{0.2em} 110 \degree \hspace{0.2em} 110° . Solve the triangle for the missing side and angles.

Again, we start with a rough sketch for the problem.

Step 1.  Use the Cosine Rule to find the missing side

When we have two sides and the included angle, the law of cosines allows us to find the third side.

Applying the cosine rule for side b \hspace{0.2em} b \hspace{0.2em} b ,

Taking square root on both sides,

Step 2.  Use the Sine Rule for the remaining angles

At this point, we know three sides and one angle. So, we can find the unknown angles using the sine rule.

When using the sine rule to find an angle, go for the smaller angle (the one opposite the smaller side) if possible. Here’s why.

The smaller angle cannot be obtuse and so you can confidently accept the acute angle solution when taking the inverse of sine.

Here though, we already have an obtuse angle in the triangle. So, no matter which of the unknown angles we calculate first, we can safely ignore the obtuse options when taking sine inverse.

Alright, now lets split the equation and solve for C \hspace{0.2em} C \hspace{0.2em} C .

And finally, as we did in the last example, let's calculate the last angle using the angle sum property.

When it comes to problems of type SSA, it’s useful to divide them into two sub-types. Let me explain with an example of each type.

Solve the triangle below for the missing sides and angles.

In this example, we don’t need to draw anything as the question itself provides the figure.

Now the question gives us two sides and one angle (duh!) but —

The known angle is adjacent to the smaller of the two known sides and hence opposite to the larger side. That’s what makes this sub-type SsA (lowercase/small “s” is placed adjacent to “A”).

Step 1.  Use the Sine Rule to find the missing angle opposite to one of the known sides

Here, we know the sides b \hspace{0.2em} b \hspace{0.2em} b and c \hspace{0.2em} c \hspace{0.2em} c and the angle B \hspace{0.2em} B \hspace{0.2em} B . So we need to find angle C \hspace{0.2em} C \hspace{0.2em} C .

Applying the sine rule to B \hspace{0.2em} B \hspace{0.2em} B and C \hspace{0.2em} C \hspace{0.2em} C .

We know C \hspace{0.2em} C \hspace{0.2em} C is acute because the opposite side, c \hspace{0.2em} c \hspace{0.2em} c , is certainly not the largest ( c < b ) \hspace{0.2em} (c < b) \hspace{0.2em} ( c < b ) .

Step 2.  Find the third angle using the angle sum property

Step 3.  Find the missing side using the sine rule

That’s it.

In △ A B C \hspace{0.2em} \triangle ABC \hspace{0.2em} △ A BC , b = 15 \hspace{0.2em} b = 15 \hspace{0.2em} b = 15 , c = 21 \hspace{0.2em} c = 21 \hspace{0.2em} c = 21 , and B = 42 ° \hspace{0.2em} B = 42 \degree \hspace{0.2em} B = 42° . Solve the triangle.

Here’s the drawing to get started.

This is an example of the other sub-type where the known angle is adjacent to the larger of the two known sides and hence opposite to the smaller side. That’s what makes this sub-type sSA (uppercase “S” is placed adjacent to “A”).

This case is often termed as the “ ambiguous case ” because, as you’ll see, it has two potential solutions. And unless we have more information about the problem, we can’t pick one over the other. So both solutions must be worked out.

Unlike in the last example, here, we need to consider both acute and obtuse values for C \hspace{0.2em} C C . That’s because side c     ( A B ) \hspace{0.2em} c \, \, (AB) \hspace{0.2em} c ( A B ) could be the largest side. It is larger than b \hspace{0.2em} b \hspace{0.2em} b , and we don’t know a \hspace{0.2em} a \hspace{0.2em} a yet.

Acute Solution for C \hspace{0.2em} C \hspace{0.2em} C

Obtuse Solution for C \hspace{0.2em} C \hspace{0.2em} C

Note — When using the sine rule, you can get the obtuse solution by subtracting the acute solution from 180 ° \hspace{0.2em} 180 \degree 180° .

Next, we'll need to work out the remainder of the solution for each of the two cases.

First, let's go ahead assuming C \hspace{0.2em} C \hspace{0.2em} C is acute.

Step 4.  Repeat steps 2 and 3 assuming C \hspace{0.2em} C \hspace{0.2em} C is obtuse.

Again, first, we find angle A \hspace{0.2em} A \hspace{0.2em} A .

And then side a \hspace{0.2em} a \hspace{0.2em} a

Great, we have found all the missing values for both possible solutions. To make things clearer, here are their drawings.

Note - Before we move ahead, I want to make it clear you don’t always have to worry about multiple solutions. Of the six cases we discuss in this tutorial, this is the only one of that kind.

Solve the triangle below.

Step 1.  Find the third angle using the angle sum property

Step 2.  Find the missing side using the sine rule

You know what to do now — split it into two equations and solve for b \hspace{0.2em} b \hspace{0.2em} b and c \hspace{0.2em} c \hspace{0.2em} c .

Solving for b \hspace{0.2em} b \hspace{0.2em} b

Solving for c \hspace{0.2em} c \hspace{0.2em} c

Triangle solved.

In △ A B C \hspace{0.2em} \triangle ABC \hspace{0.2em} △ A BC , c = 14 \hspace{0.2em} c = 14 \hspace{0.2em} c = 14 , B = 45 ° \hspace{0.2em} B = 45 \degree \hspace{0.2em} B = 45° , and C = 55 ° \hspace{0.2em} C = 55 \degree \hspace{0.2em} C = 55° . Solve the triangle.

Step 1.  Here, we know two angles. So, we find the third angle using the angle sum property

Step 2.  Find the missing sides using the sine rule

Solving for a \hspace{0.2em} a \hspace{0.2em} a

And with that, we come to the end of this tutorial on solving triangles. Until next time.

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A triangle is a type of polygon .

• 1 Definition
• 2.1 Equilateral
• 2.2 Isosceles
• 2.3 Scalene
• 3 Related Formulas and Theorems
• 4 External Links

A triangle is any polygon with three sides , with the smaller angle measures of the intersections of the sides summing to 180 degrees. Triangles exist in Euclidean geometry , and are the simplest possible polygon. In physics , triangles are noted for their durability, since they have only three vertices around with to distort.

Triangles are split into six categories; three by their angles and three by their side lengths.

Equilateral

An isosceles triangle has at least two congruent sides (this means that all equilateral triangles are also isosceles), and the two angles opposite the congruent sides are also congruent (this is commonly known as the Hinge theorem ).

A scalene triangle has no congruent sides, and also no congruent angles (this can be proven by the converse of the Hinge Theorem).

A right triangle has a right angle , which means the other two angles are complementary . Trigonometry is largely based on right triangles, and the famous Pythagorean Theorem deals with the side lengths of the right triangle. Note that it is impossible to have an equilateral right triangle. It is also impossible to have more than one right angle in a triangle.

An obtuse triangle has an obtuse angle . Note that it is impossible to have an equilateral obtuse triangle. It is also impossible to have more than one obtuse angle in a triangle.

All the angles of an acute triangle are acute angles .

Related Formulas and Theorems

• In any triangle, the sum of any two sides is greater than the length of the third side.

• See trigonometric identities for a list of formulae related to trigonometry.

External Links

• Introduction to Geometry by Richard Rusczyk
• Challenging Problems in Geometry - A good book for students who already have a solid handle on elementary geometry.
• Geometry Revisited - A classic.
• Mathematical Problems - Lecture delivered before the International Congress of Mathematicians at Paris in 1900 by David Hilbert.
• Foundations of Geometry by David Hilbert
• Circumcircle

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Tri-Triangles

In the problem Tri-Triangles , students use algebraic thinking to solve problems involving patterns, sequences, generalizations, and linear and non-linear functions. The mathematical topics that underlie this problem are finding and extending patterns, creating generalizations, finding functions, developing inverse processes, exploring non-linear functions, and justifying solutions. In each level, students must make sense of the problem and persevere in solving it ( MP.1 ). Each problem is divided into five levels of difficulty, Level A through Level E, to allow access and scaffolding for students into different aspects of the problem and to stretch students to go deeper into mathematical complexity.

PRE-K In this task, students are shown a triangle pattern and use toothpicks to extend the pattern and identify the number of toothpicks that would be needed in subsequent patterns.

LEVEL A      In this level, students view a pattern of triangles composed of toothpicks.  Their task is to determine the number of toothpicks that make up each pattern. 

In this level, students use multiplication and division within 100 to solve word problems in situations involving equal groups ( 3.OA.A.3 ).          LEVEL B      In this level, students examine a linear pattern that involves a constant. The task involves a set of triangular tables that are arranged adjacently in a row. The task asks students to determine the relationship between the number of tables and the number of people who can sit around the tables.  Students need to extend the pattern. They also find the inverse relationship, i.e., find the pattern number when given the total number of people seated. 

In this level, students are given a shape pattern and asked to extend the pattern and informally describe the rule of the pattern ( 4.OA.C.5 ).  

LEVEL C      In this level, students determine how a pattern grows. Students need to see that the pattern grows by square numbers. They identify the relationship and then explain a valid process for finding these values. 

In this level, students  determine how a pattern of triangles grows. Students use their understanding of expressions and exponents ( 6.EE.A.1 ) to identify the relationship between the pattern and number of triangles and then explain a valid process for finding these values ( 6.EE.B.6, 6.EE.C.9 ).          LEVEL D      In this level, students are asked to generalize a rule for finding a value in the triangular number sequence. They are also asked to explain the process for finding an inverse value for the triangular number sequence by finding the term when given the total.

Students explore a pattern of triangular numbers and write an explicit expression describing the pattern ( F-BF.A.1a ). Students use the expression to write a quadratic equation to find the pattern number given the triangular number ( A-CED.A.1 ). The quadratic equation can be solved with reasoning, factoring, or the quadratic formula ( A-REI.B.4b ). Students explain their reasoning ( SMP.3 ) and use visual models to explore the pattern ( SMP.5 ).          LEVEL E      In this level, students generate a closed expression for a sequence that grows as the sum of two exponential functions.  In addition, the students must justify their findings. This level builds on the work students do in level D. 

In this level, students build on their experience with triangular numbers from level D to generate an explicit expression to describe a pattern ( F-BF.A.1a ). As they build the explicit expression, they interpret parts of the expression independently and then combine the parts to arrive at the final expression ( A-SSE.A.2 ). The rule that is developed may use exponential or polynomial functions. This level requires students to look for and make use of structure in the expressions they build ( SMP.7 ).

PROBLEM OF THE MONTH  Download the complete packet of Tri-Triangles Levels A-E here . 

You can learn more about how to implement these problems in a school-wide Problem of the Month initiative in “Jumpstarting a Schoolwide Culture of Mathematical Thinking: Problems of the Month,” a practitioner’s guide.   Download the guide as iBook with embedded videos or Download as PDF without embedded videos .

SOLUTIONS To request the Inside Problem Solving Solutions Guide, please get in touch with us via the feedback form .

Introduction to Trigonometry

Trigonometry (from Greek trigonon "triangle" + metron "measure")

Want to learn Trigonometry? Here is a quick summary. Follow the links for more, or go to Trigonometry Index

Trigonometry helps us find angles and distances, and is used a lot in science, engineering, video games, and more!

Right-Angled Triangle

The triangle of most interest is the right-angled triangle . The right angle is shown by the little box in the corner:

Another angle is often labeled θ , and the three sides are then called:

• Adjacent : adjacent (next to) the angle θ
• Opposite : opposite the angle θ
• and the longest side is the Hypotenuse

Why a Right-Angled Triangle?

Why is this triangle so important?

Imagine we can measure along and up but want to know the direct distance and angle:

Trigonometry can find that missing angle and distance.

Or maybe we have a distance and angle and need to "plot the dot" along and up:

Questions like these are common in engineering, computer animation and more.

And trigonometry gives the answers!

Sine, Cosine and Tangent

The main functions in trigonometry are Sine, Cosine and Tangent

They are simply one side of a right-angled triangle divided by another.

For any angle " θ ":

(Sine, Cosine and Tangent are often abbreviated to sin, cos and tan .)

Example: What is the sine of 35°?

Using this triangle (lengths are only to one decimal place):

sin(35°) = Opposite Hypotenuse = 2.8 4.9 = 0.57...

The triangle could be larger, smaller or turned around, but that angle will always have that ratio .

Calculators have sin, cos and tan to help us, so let's see how to use them:

Example: How Tall is The Tree?

We can't reach the top of the tree, so we walk away and measure an angle (using a protractor) and distance (using a laser):

• We know the Hypotenuse
• And we want to know the Opposite

Sine is the ratio of Opposite / Hypotenuse :

sin(45°) = Opposite Hypotenuse

Get a calculator, type in "45", then the "sin" key:

sin(45°) = 0.7071...

What does the 0.7071... mean? It is the ratio of the side lengths, so the Opposite is about 0.7071 times as long as the Hypotenuse.

We can now put 0.7071... in place of sin(45°):

0.7071... = Opposite Hypotenuse

And we also know the hypotenuse is 20 :

0.7071... = Opposite 20

To solve, first multiply both sides by 20:

20 × 0.7071... = Opposite

Opposite = 14.14m (to 2 decimals)

The tree is 14.14m tall

Try Sin Cos and Tan

Play with this for a while (move the mouse around) and get familiar with values of sine, cosine and tangent for different angles, such as 0°, 30°, 45°, 60° and 90°.

Also try 120°, 135°, 180°, 240°, 270° etc, and notice that positions can be positive or negative by the rules of Cartesian coordinates , so the sine, cosine and tangent change between positive and negative also.

So trigonometry is also about circles !

Unit Circle

What you just played with is the Unit Circle .

It is a circle with a radius of 1 with its center at 0.

Because the radius is 1, we can directly measure sine, cosine and tangent.

Here we see the sine function being made by the unit circle:

Note: you can see the nice graphs made by sine, cosine and tangent .

Degrees and Radians

Angles can be in Degrees or Radians . Here are some examples:

Repeating Pattern

Because the angle is rotating around and around the circle the Sine, Cosine and Tangent functions repeat once every full rotation (see Amplitude, Period, Phase Shift and Frequency ).

When we want to calculate the function for an angle larger than a full rotation of 360° (2 π radians) we subtract as many full rotations as needed to bring it back below 360° (2 π radians):

Example: what is the cosine of 370°?

370° is greater than 360° so let us subtract 360°

370° − 360° = 10°

cos(370°) = cos(10°) = 0.985 (to 3 decimal places)

And when the angle is less than zero, just add full rotations.

Example: what is the sine of −3 radians?

−3 is less than 0 so let us add 2 π radians

−3 + 2 π = −3 + 6.283... = 3.283... rad ians

sin(−3) = sin(3.283...) = −0.141 (to 3 decimal places)

Solving Triangles

Trigonometry is also useful for general triangles, not just right-angled ones .

It helps us in Solving Triangles . "Solving" means finding missing sides and angles.

Example: Find the Missing Angle "C"

Angle C can be found using angles of a triangle add to 180° :

So C = 180° − 76° − 34° = 70°

We can also find missing side lengths. The general rule is:

When we know any 3 of the sides or angles we can find the other 3 (except for the three angles case)

See Solving Triangles for more details.

Other Functions (Cotangent, Secant, Cosecant)

Similar to Sine, Cosine and Tangent, there are three other trigonometric functions which are made by dividing one side by another:

Trigonometric and Triangle Identities

And as you get better at Trigonometry you can learn these:

Enjoy becoming a triangle (and circle) expert!

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‘Groups’ Underpin Modern Math. Here’s How They Work.

September 6, 2024

Adam Nickel for  Quanta Magazine

Introduction

Mathematics started with numbers — clear, concrete, intuitive. Over the last two centuries, however, it has become a far more abstract enterprise. One of the first major steps down this road was taken in the late 18th and early 19th centuries. It involved a field called group theory, and it changed math — theoretical and applied — as we know it.

Groups generalize essential properties of the whole numbers. They have transformed geometry, algebra and analysis, the mathematical study of smoothly changing functions. They’re used to encrypt messages and study the shapes of viruses . Physicists rely on them to unify the fundamental forces of nature: At high energies, group theory can be used to show that electromagnetism and the forces that hold atomic nuclei together and cause radioactivity are all manifestations of a single underlying force.

The term “group” in a mathematical context was coined in 1830 by Évariste Galois, a French prodigy, just 18 years old at the time. (Two years later, he would be killed in a duel, having already changed the course of mathematical history.) But he didn’t discover groups single-handed. “It’s not like a bunch of mathematicians got together one day and said, ‘Let’s create an abstract structure just for a laugh,’” said Sarah Hart, a group theorist at Gresham College in London. “It emerged gradually, over maybe 50 years in the 19th century, that these were the right rules to require. They give you the most flexibility and generality, while still allowing you to prove things.”

Évariste Galois helped lay the foundations for group theory as a teenager.

Public domain

A group is a set, or collection of objects, together with an operation that takes in two objects and outputs a third. Arguably the simplest example is the integers and the operation of addition. Groups must satisfy four rules.

• The first is called closure: Add any two integers and you’ll get another integer.
• The second rule is called associativity: If you add three numbers together, the result doesn’t depend on how you group them. You can add 3 and 4 to get 7, then add 5 to get 12. Or you can add 3 to the sum of 4 and 5. Either way, you’ll still get the same answer: 12 = (3 + 4) + 5 = 3 + (4 + 5).
• The third rule is that the group must contain an element that leaves other group elements unchanged, called an identity element. The number zero is the identity for addition, since adding zero to a number keeps that number the same.
• Finally, each group element must have an inverse — add an element and its inverse, and you’ll get the identity. In the integers, the inverse of a number is its negative. For example, 3 + (−3) = 0.

To understand the significance of these four properties, it helps to consider a noteworthy omission. When adding two numbers together, you can change the order without affecting the outcome: 3 + 5 is the same as 5 + 3. This property is called commutativity. But there’s no requirement that groups be commutative. By making this property optional, mathematicians have been able to explore a rich variety of structures.

For an example of a noncommutative group, consider an equilateral triangle with labeled corners. If you rotate the triangle a third of the way around or flip it along its vertical axis, the only thing that will change about the image is the locations of the labels. There are six of these transformations that leave the shape otherwise unchanged, called symmetries of the triangle. They form a group called D 6 . (More generally, D 2 n is a group formed by the symmetries of a regular shape with n sides, so D 8 is the group of symmetries of a square.)

5W Infographics/Mark Belan for Quanta Magazine

To “add” two symmetries, just do one after another. You will quickly find that D 6 is not commutative: Flipping then rotating leaves the labels in different places than they will be if you rotate then flip.

D 6 is one of only two possible groups with six elements. For the other example of a six-element group, take the numbers {0, 1, 2, 3, 4, 5} as the set. For the operation, add two numbers in the usual way and then divide by 6, ignoring the quotient but retaining the remainder. So 3 and 5 deliver 2, since 8 leaves a remainder of 2 when divided by 6. This is called addition modulo 6, and the group is called Z 6 . In general, Z n is a group with n elements obtained from the numbers {0, 1, 2, 3, …,  n − 1} together with addition modulo n . Unlike D 6 , Z 6 is commutative, because 3 + 5 = 5 + 3, and so forth.

Z 6 and D 6 have different structures. Not only is one commutative and the other not, but you can generate any element of Z 6 using just one of its elements, the number 1: Start with 1, then keep adding 1. In D 6 , no element has this property. Figuring out the possible structures of groups has been one of the central projects of algebra over the last century.

To do this, mathematicians try to identify smaller groups contained within a group, called subgroups. These must retain the operation used for the full group. For example, the even integers form a subgroup within the integers. An even integer plus an even integer always results in another even integer. On the other hand, the odd numbers are not a subgroup, since if you add two odd numbers, you’ll get an even number. The identity element always forms a subgroup by itself, called the trivial subgroup.

Figuring out what subgroups a group contains is one way to understand its structure. For example, the subgroups of Z 6 are {0}, {0, 2, 4} and {0, 3} — the trivial subgroup, the multiples of 2, and the multiples of 3. In the group D 6 , rotations form a subgroup, but reflections don’t. That’s because two reflections performed in sequence produce a rotation, not a reflection, just as adding two odd numbers results in an even one.

Certain types of subgroups called “normal” subgroups are especially helpful to mathematicians. In a commutative group, all subgroups are normal, but this isn’t always true more generally. These subgroups retain some of the most useful properties of commutativity, without forcing the entire group to be commutative. If a list of normal subgroups can be identified, groups can be broken up into components much the way integers can be broken up into products of primes. Groups that have no normal subgroups are called simple groups and cannot be broken down any further, just as prime numbers can’t be factored. The group Z n is simple only when n is prime — the multiples of 2 and 3, for instance, form normal subgroups in Z 6 .

However, simple groups are not always so simple. “It’s the biggest misnomer in mathematics,” Hart said. In 1892, the mathematician Otto Hölder proposed that researchers assemble a complete list of all possible finite simple groups. (Infinite groups such as the integers form their own field of study.)

It turns out that almost all finite simple groups either look like Z n (for prime values of n ) or fall into one of two other families. And there are 26 exceptions, called sporadic groups. Pinning them down, and showing that there are no other possibilities, took over a century.

The largest sporadic group, aptly called the monster group, was discovered in 1973. It has more than 8 × 10 54 elements and represents geometric rotations in a space with nearly 200,000 dimensions. “It’s just crazy that this thing could be found by humans,” Hart said.

By the 1980s, the bulk of the work Hölder had called for appeared to have been completed, but it was tough to show that there were no more sporadic groups lingering out there. The classification was further delayed when, in 1989, the community found gaps in one 800-page proof from the early 1980s. A new proof was finally published in 2004, finishing off the classification.

Many structures in modern math — rings, fields and vector spaces, for example — are created when more structure is added to groups. In rings, you can multiply as well as add and subtract; in fields, you can also divide. But underneath all of these more intricate structures is that same original group idea, with its four axioms. “The richness that’s possible within this structure, with these four rules, is mind-blowing,” Hart said.

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A framework for solving parabolic partial differential equations

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Computer graphics and geometry processing research provide the tools needed to simulate physical phenomena like fire and flames, aiding the creation of visual effects in video games and movies as well as the fabrication of complex geometric shapes using tools like 3D printing.

Under the hood, mathematical problems called partial differential equations (PDEs) model these natural processes. Among the many PDEs used in physics and computer graphics, a class called second-order parabolic PDEs explain how phenomena can become smooth over time. The most famous example in this class is the heat equation, which predicts how heat diffuses along a surface or in a volume over time.

Researchers in geometry processing have designed numerous algorithms to solve these problems on curved surfaces, but their methods often apply only to linear problems or to a single PDE. A more general approach by researchers from MIT’s Computer Science and Artificial Intelligence Laboratory (CSAIL) tackles a general class of these potentially nonlinear problems.  In a paper recently published in the Transactions on Graphics journal and presented at the SIGGRAPH conference, they describe an algorithm that solves different nonlinear parabolic PDEs on triangle meshes by splitting them into three simpler equations that can be solved with techniques graphics researchers already have in their software toolkit. This framework can help better analyze shapes and model complex dynamical processes.

“We provide a recipe: If you want to numerically solve a second-order parabolic PDE, you can follow a set of three steps,” says lead author Leticia Mattos Da Silva SM ’23, an MIT PhD student in electrical engineering and computer science (EECS) and CSAIL affiliate. “For each of the steps in this approach, you’re solving a simpler problem using simpler tools from geometry processing, but at the end, you get a solution to the more challenging second-order parabolic PDE.” To accomplish this, Da Silva and her coauthors used Strang splitting, a technique that allows geometry processing researchers to break the PDE down into problems they know how to solve efficiently.

First, their algorithm advances a solution forward in time by solving the heat equation (also called the “diffusion equation”), which models how heat from a source spreads over a shape. Picture using a blow torch to warm up a metal plate — this equation describes how heat from that spot would diffuse over it. 
This step can be completed easily with linear algebra.

Now, imagine that the parabolic PDE has additional nonlinear behaviors that are not described by the spread of heat. This is where the second step of the algorithm comes in: it accounts for the nonlinear piece by solving a Hamilton-Jacobi (HJ) equation, a first-order nonlinear PDE.  While generic HJ equations can be hard to solve, Mattos Da Silva and coauthors prove that their splitting method applied to many important PDEs yields an HJ equation that can be solved via convex optimization algorithms. Convex optimization is a standard tool for which researchers in geometry processing already have efficient and reliable software. In the final step, the algorithm advances a solution forward in time using the heat equation again to advance the more complex second-order parabolic PDE forward in time.


Among other applications, the framework could help simulate fire and flames more efficiently. “There’s a huge pipeline that creates a video with flames being simulated, but at the heart of it is a PDE solver,” says Mattos Da Silva. For these pipelines, an essential step is solving the G-equation, a nonlinear parabolic PDE that models the front propagation of the flame and can be solved using the researchers’ framework.

The team’s algorithm can also solve the diffusion equation in the logarithmic domain, where it becomes nonlinear. Senior author Justin Solomon, associate professor of EECS and leader of the CSAIL Geometric Data Processing Group, previously developed a state-of-the-art technique for optimal transport that requires taking the logarithm of the result of heat diffusion. Mattos Da Silva’s framework provided more reliable computations by doing diffusion directly in the logarithmic domain. This enabled a more stable way to, for example, find a geometric notion of average among distributions on surface meshes like a model of a koala. Even though their framework focuses on general, nonlinear problems, it can also be used to solve linear PDE. For instance, the method solves the Fokker-Planck equation, where heat diffuses in a linear way, but there are additional terms that drift in the same direction heat is spreading. In a straightforward application, the approach modeled how swirls would evolve over the surface of a triangulated sphere. The result resembles purple-and-brown latte art.

The researchers note that this project is a starting point for tackling the nonlinearity in other PDEs that appear in graphics and geometry processing head-on. For example, they focused on static surfaces but would like to apply their work to moving ones, too. Moreover, their framework solves problems involving a single parabolic PDE, but the team would also like to tackle problems involving coupled parabolic PDE. These types of problems arise in biology and chemistry, where the equation describing the evolution of each agent in a mixture, for example, is linked to the others’ equations.

Mattos Da Silva and Solomon wrote the paper with Oded Stein, assistant professor at the University of Southern California’s Viterbi School of Engineering. Their work was supported, in part, by an MIT Schwarzman College of Computing Fellowship funded by Google, a MathWorks Fellowship, the Swiss National Science Foundation, the U.S. Army Research Office, the U.S. Air Force Office of Scientific Research, the U.S. National Science Foundation, MIT-IBM Watson AI Lab, the Toyota-CSAIL Joint Research Center, Adobe Systems, and Google Research.

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