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Difference between revisions of "Pythagorean Theorem"

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Since <math>ABC</math>, <math>CBH</math>, <math>ACH</math> are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths,
 
Since <math>ABC</math>, <math>CBH</math>, <math>ACH</math> are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths,
 
<cmath>\frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2}.</cmath>
 
<cmath>\frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2}.</cmath>
−
But since triangle <math>ABC</math> is composed of triangles <math>CBH</math> and <math>ACH</math>, <math>[ABC] = [CBH]+[ACH]</math>, so <math>AB^2 = CB^2 + AC^2</math>. {{Halmos}}
+
But since triangle <math>ABC</math> is composed of triangles <math>CBH</math> and <math>ACH</math>, <math>[ABC] = [CBH]+[ACH]</math>, so <math>AB^2 = CB^2 + AC^2</math>.
  
 
=== Proof 3 ===
 
=== Proof 3 ===
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Evidently, <math>AY = AB-BC</math> and <math>AX = AB+BC</math>.  By considering the [[Power of a Point]] <math>A</math> with respect to <math>\omega</math>, we see
 
Evidently, <math>AY = AB-BC</math> and <math>AX = AB+BC</math>.  By considering the [[Power of a Point]] <math>A</math> with respect to <math>\omega</math>, we see
−
<cmath>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2.</cmath> {{Halmos}}
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<cmath>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2.</cmath>
  
 
=== Proof 4 ===
 
=== Proof 4 ===
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</asy>
 
</asy>
 
</center>
 
</center>
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<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}}
+
<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>.
  
 
== Pythagorean Triples ==
 
== Pythagorean Triples ==
  
 
{{main|Pythagorean triple}}
 
{{main|Pythagorean triple}}
−
A [[Pythagorean triple]] is a [[triple]] of [[positive integer]]s such that <math>a^{2}+b^{2}=c^{2}</math>. All such triples contain numbers which are side lengths of the sides of a right triangle.  Among these, the [[Primitive Pythagorean triple]]s, are those in which the three numbers are [[coprime]].  A few of them are:
+
A [[Pythagorean triple]] is a of [[positive integer]]s such that <math>a^{2}+b^{2}=c^{2}</math>. All such triples contain numbers which are side lengths of the sides of a right triangle.  Among these, the [[Primitive Pythagorean triple]]s, are those in which the three numbers are [[coprime]].  A few of them are:
  
 
<cmath>3-4-5</cmath>
 
<cmath>3-4-5</cmath>
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<cmath>21-72-75 = (7-24-25)*3</cmath>
 
<cmath>21-72-75 = (7-24-25)*3</cmath>
 
<cmath>10-24-26 = (5-12-13)*2</cmath>
 
<cmath>10-24-26 = (5-12-13)*2</cmath>
 +
 +
Also note that one easy way to find Pythagorean triples is as follows.  Choose any odd number <math>n</math>. Find <math>n^2</math>. Find <math>\frac{n-1}{2}</math> and <math>\frac{n-1}{2} + 1</math>. Your Pythagorean triple is <math>n</math>, <math>\frac{n-1}{2}</math>, and <math>\frac{n-1}{2} + 1</math>.
 +
 +
~ [[User:Liu7113|liu7113]]
  
 
== Problems ==
 
== Problems ==
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<math>\sqrt{333^2+444^2} = \boxed{555}</math>.
 
<math>\sqrt{333^2+444^2} = \boxed{555}</math>.
 
==== Solution 2 (Using 3-4-5) ====
 
==== Solution 2 (Using 3-4-5) ====
−
<math>333-444-555</math> is the resulting Pythagorean triple when <math>3-4-5</math> is multiplied by <math>11</math>, so the answer is <math>\boxed{555}</math>.
+
<math>333-444-555</math> is the resulting Pythagorean triple when <math>3-4-5</math> is multiplied by <math>111</math>, so the answer is <math>\boxed{555}</math>.
  
 
=== Problem 2 ===
 
=== Problem 2 ===
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There are no more cases as the hypotenuse has to be greater than the leg, so the sum is <math>4+5=\boxed{9}</math>.
 
There are no more cases as the hypotenuse has to be greater than the leg, so the sum is <math>4+5=\boxed{9}</math>.
  
 +
==Videos==
 +
===Video by YippieMath(Quick and Easy)===
 +
https://www.youtube.com/watch?v=JMZvZtvkoME
 
== External links ==
 
== External links ==
 
*[http://www.cut-the-knot.org/pythagoras/index.shtml 122 proofs of the Pythagorean Theorem]
 
*[http://www.cut-the-knot.org/pythagoras/index.shtml 122 proofs of the Pythagorean Theorem]

Latest revision as of 17:37, 17 May 2025

The Pythagorean Theorem states that for a right triangle with legs of length $a$ and $b$ and hypotenuse of length $c$ we have the relationship $a^2+b^2=c^2$. This theorem has been known since antiquity and is a classic to prove; hundreds of proofs have been published and many can be demonstrated entirely visually. 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.

Proofs

In these proofs, we will let $ABC$ be any right triangle with a right angle at $\angle ACB$, and we use $[ABC]$ to denote the area of triangle $ABC$.

Proof 1

Let $D$ be the foot of the altitude from $C$. $ABC$, $ACD$, $BCD$ are similar triangles, so $\frac{AC}{AD}=\frac{AB}{AC} \implies AC^2=(AD)(AB)$ and $\frac{BC}{BD}=\frac{AB}{BC} \implies BC^2=(BD)(AB)$. Adding these equations gives us $AC^2+BC^2=(AD)(AB)+(BD)(AB) \implies AC^2+BC^2=(AB)(AD+BD) \implies AC^2+BC^2=AB^2$

Proof 2

Let $H$ be the foot of the altitude from $C$.

[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]

Since $ABC$, $CBH$, $ACH$ are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, \[\frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2}.\] But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH]+[ACH]$, so $AB^2 = CB^2 + AC^2$.

Proof 3

Consider a circle $\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\omega$. Let the line $AB$ meet $\omega$ at $Y$ and $X$, as shown in the diagram:

Pyth2.png

Evidently, $AY = AB-BC$ and $AX = AB+BC$. By considering the Power of a Point $A$ with respect to $\omega$, we see \[AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2.\]

Proof 4

$ABCD$ and $EFGH$ are squares.

[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]

$(a+b)^2=c^2+4\left(\frac{1}{2}ab\right)\implies a^2+2ab+b^2=c^2+2ab\implies a^2+b^2=c^2$.

Pythagorean Triples

Main article: Pythagorean triple

A Pythagorean triple is a of positive integers such that $a^{2}+b^{2}=c^{2}$. All such triples contain numbers which are side lengths of the sides of a right triangle. Among these, the Primitive Pythagorean triples, are those in which the three numbers are coprime. A few of them are:

\[3-4-5\] \[5-12-13\] \[7-24-25\] \[8-15-17\] \[9-40-41\] \[12-35-37\] \[20-21-29\] \[11-60-61\]

Note that (3,4,5) is the only Pythagorean triple that consists of consecutive integers.

Any triple created by multiplying all three numbers in a Pythagorean triple by a positive integer is Pythagorean. In other words, if (a,b,c) is a Pythagorean triple it follows that (ka,kb,kc) will also form a Pythagorean triple for any positive integer constant k. For example, \[6-8-10 = (3-4-5)*2\] \[21-72-75 = (7-24-25)*3\] \[10-24-26 = (5-12-13)*2\]

Also note that one easy way to find Pythagorean triples is as follows. Choose any odd number $n$. Find $n^2$. Find $\frac{n-1}{2}$ and $\frac{n-1}{2} + 1$. Your Pythagorean triple is $n$, $\frac{n-1}{2}$, and $\frac{n-1}{2} + 1$.

~ liu7113

Problems

Introductory

Problem 1

Right triangle $ABC$ has legs of length $333$ and $444$. Find the hypotenuse of $ABC$.

Solution 1 (Bash)

$\sqrt{333^2+444^2} = \boxed{555}$.

Solution 2 (Using 3-4-5)

$333-444-555$ is the resulting Pythagorean triple when $3-4-5$ is multiplied by $111$, so the answer is $\boxed{555}$.

Problem 2

Right triangle $ABC$ has side lengths of $3$ and $4$. 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, so the sum is $4+5=\boxed{9}$.

Videos

Video by YippieMath(Quick and Easy)

https://www.youtube.com/watch?v=JMZvZtvkoME

External links

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