# Difference between revisions of "Pythagorean Theorem"

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Let <math>H </math> be the perpendicular to side <math>AB </math> from <math>{} C </math>. | Let <math>H </math> be the perpendicular to side <math>AB </math> from <math>{} C </math>. | ||

− | <center> | + | <center> |

+ | <asy> | ||

+ | pair A, B, C, H; | ||

+ | A = (0, 0); | ||

+ | B = (4, 3); | ||

+ | C = (4, 0); | ||

+ | H = foot(C, A, B); | ||

+ | |||

+ | draw(A--B--C--cycle); | ||

+ | draw(C--H); | ||

+ | draw(rightanglemark(A, C, B)); | ||

+ | draw(rightanglemark(C, H, B)); | ||

+ | label("$A$", A, SSW); | ||

+ | label("$B$", B, ENE); | ||

+ | label("$C$", C, SE); | ||

+ | label("$H$", H, NNW); | ||

+ | </asy> | ||

+ | </center> | ||

− | Since <math>ABC, CBH, ACH </math> are similar right triangles, and the areas of similar triangles are | + | Since <math>ABC, CBH, ACH</math> are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, |

<center> | <center> | ||

<math> \frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2} </math>. | <math> \frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2} </math>. | ||

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<center>[[Image:Pyth2.png]]</center> | <center>[[Image:Pyth2.png]]</center> | ||

− | Evidently, <math>AY = AB - BC </math> and <math>AX = AB + BC </math>. By considering the [[ | + | Evidently, <math>AY = AB - BC </math> and <math>AX = AB + BC </math>. By considering the [[Power of a Point | power of point]] <math>A </math> with respect to <math>\omega </math>, we see |

<center> | <center> | ||

<math>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 </math>. {{Halmos}} | <math>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 </math>. {{Halmos}} | ||

</center> | </center> | ||

+ | |||

+ | === Proof 3 === | ||

+ | |||

+ | <math>ABCD</math> and <math>EFGH</math> are squares. | ||

+ | <center> | ||

+ | <asy> | ||

+ | pair A, B,C,D; | ||

+ | A = (-10,10); | ||

+ | B = (10,10); | ||

+ | C = (10,-10); | ||

+ | D = (-10,-10); | ||

+ | |||

+ | pair E,F,G,H; | ||

+ | E = (7,10); | ||

+ | F = (10, -7); | ||

+ | G = (-7, -10); | ||

+ | H = (-10, 7); | ||

+ | |||

+ | draw(A--B--C--D--cycle); | ||

+ | label("$A$", A, NNW); | ||

+ | label("$B$", B, ENE); | ||

+ | label("$C$", C, ESE); | ||

+ | label("$D$", D, SSW); | ||

+ | |||

+ | draw(E--F--G--H--cycle); | ||

+ | label("$E$", E, N); | ||

+ | label("$F$", F,SE); | ||

+ | label("$G$", G, S); | ||

+ | label("$H$", H, W); | ||

+ | |||

+ | label("a", A--B,N); | ||

+ | label("a", B--F,SE); | ||

+ | label("a", C--G,S); | ||

+ | label("a", H--D,W); | ||

+ | label("b", E--B,N); | ||

+ | label("b", F--C,SE); | ||

+ | label("b", G--D,S); | ||

+ | label("b", A--H,W); | ||

+ | label("c", E--H,NW); | ||

+ | label("c", E--F); | ||

+ | label("c", F--G,SE); | ||

+ | label("c", G--H,SW); | ||

+ | </asy> | ||

+ | </center> | ||

+ | <math>(a+b)^2=c^2+4\left(\frac{1}{2}ab\right)\implies a^2+2ab+b^2=c^2+2ab\implies a^2 + b^2=c^2</math>. {{Halmos}} | ||

== Common Pythagorean Triples == | == Common Pythagorean Triples == | ||

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<cmath>8-15-17</cmath> | <cmath>8-15-17</cmath> | ||

<cmath>9-40-41</cmath> | <cmath>9-40-41</cmath> | ||

+ | <cmath>12-35-37</cmath> | ||

<cmath>20-21-29</cmath> | <cmath>20-21-29</cmath> | ||

+ | <cmath>11-60-61</cmath> | ||

+ | |||

+ | |||

+ | Also Pythagorean Triples can be created with the a Pythagorean triple by multiplying the lengths by any integer. | ||

+ | For example, | ||

+ | <cmath>6-8-10 = (3-4-5)*2</cmath> | ||

+ | <cmath>21-72-75 = (7-24-25)*3</cmath> | ||

+ | <cmath>10-24-26 = (5-12-13)*2</cmath> | ||

== Problems == | == Problems == | ||

Line 47: | Line 118: | ||

* [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]] | * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]] | ||

* [[2007 AMC 12A Problems/Problem 10 | 2007 AMC 12A Problem 10]] | * [[2007 AMC 12A Problems/Problem 10 | 2007 AMC 12A Problem 10]] | ||

+ | === Sample Problem === | ||

+ | Right triangle <math>ABC</math> has legs of length <math>333</math> and <math>444</math>. Find the hypotenuse of <math>ABC</math>. | ||

+ | ==== Solution 1 (Bash) ==== | ||

+ | <math>\sqrt{333^2 + 444^2} = 555</math>. | ||

+ | ==== Solution 2 (Using 3-4-5) ==== | ||

+ | We see <math>333-444</math> looks like the legs of a <math>3-4-5</math> right triangle with a multiplication factor of 111. Thus <math>5*111 = 555</math>. | ||

+ | |||

+ | === Another Problem === | ||

+ | Right triangle <math>ABC</math> has side lengths of <math>3</math> and <math>4</math>. Find the sum of all the possible hypotenuses. | ||

+ | ==== Solution (Casework) ==== | ||

+ | Case 1: 3 and 4 are the legs. Then 5 is the hypotenuse. | ||

+ | Case 2: 3 is a leg and 4 is the hypotenuse. | ||

+ | There are no more cases as the hypotenuse has to be greater than the leg. | ||

+ | This makes the sum <math>4+5=9</math>. | ||

== External links == | == External links == | ||

− | *[http://www.cut-the-knot.org/pythagoras/index.shtml | + | *[http://www.cut-the-knot.org/pythagoras/index.shtml 118 proofs of the Pythagorean Theorem] |

[[Category:Geometry]] | [[Category:Geometry]] | ||

Line 56: | Line 141: | ||

[[Category:Geometry]] | [[Category:Geometry]] | ||

− |

## Latest revision as of 18:38, 26 January 2020

The **Pythagorean Theorem** states that for a right triangle with legs of length and and hypotenuse of length we have the relationship . This theorem has been know since antiquity and is a classic to prove; hundreds of proofs have been published and many can be demonstrated entirely visually(the book *The Pythagorean Proposition* alone consists of more than 370). The Pythagorean Theorem is one of the most frequently used theorems in geometry, and is one of the many tools in a good geometer's arsenal. A very large number of geometry problems can be solved by building right triangles and applying the Pythagorean Theorem.

This is generalized by the Pythagorean Inequality and the Law of Cosines.

## Contents

## Proofs

In these proofs, we will let be any right triangle with a right angle at .

### Proof 1

We use to denote the area of triangle .

Let be the perpendicular to side from .

Since are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths,

.

But since triangle is composed of triangles and , , so . ∎

### Proof 2

Consider a circle with center and radius . Since and are perpendicular, is tangent to . Let the line meet at and , as shown in the diagram:

Evidently, and . By considering the power of point with respect to , we see

. ∎

### Proof 3

and are squares.

. ∎

## Common Pythagorean Triples

A Pythagorean Triple is a set of 3 positive integers such that , i.e. the 3 numbers can be the lengths of the sides of a right triangle. Among these, the Primitive Pythagorean Triples, those in which the three numbers have no common divisor, are most interesting. A few of them are:

Also Pythagorean Triples can be created with the a Pythagorean triple by multiplying the lengths by any integer.
For example,

## Problems

### Introductory

### Sample Problem

Right triangle has legs of length and . Find the hypotenuse of .

#### Solution 1 (Bash)

.

#### Solution 2 (Using 3-4-5)

We see looks like the legs of a right triangle with a multiplication factor of 111. Thus .

### Another Problem

Right triangle has side lengths of and . Find the sum of all the possible hypotenuses.

#### Solution (Casework)

Case 1: 3 and 4 are the legs. Then 5 is the hypotenuse. Case 2: 3 is a leg and 4 is the hypotenuse. There are no more cases as the hypotenuse has to be greater than the leg. This makes the sum .