Difference between revisions of "Pythagorean Theorem"

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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 know 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 $\displaystyle ABC$ be any right triangle with a right angle at $\displaystyle {} C$ .

Proof 1

We use $\displaystyle [ABC]$ to denote the area of triangle $\displaystyle ABC$ .

Let $\displaystyle H$ be the perpendicular to side $\displaystyle AB$ from $\displaystyle {} C$ .

Since $\displaystyle ABC, CBF, ACF$ are similar right triangles, and the areas of similar triangles are proportionate to the squares of corresponding side lengths,

$\frac{[ABC]}{AB^2} = \frac{[CBF]}{CB^2} = \frac{[ACF]}{AC^2}$ .

But since triangle $\displaystyle ABC$ is composed of triangles $\displaystyle CBF$ and $\displaystyle ACF$ , $\displaystyle [ABC] = [CBF] + [ACF]$ , so $\displaystyle AB^2 = CB^2 + AC^2$ . ∎

Proof 2

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

Evidently, $\displaystyle AY = AB - BC$ and $\displaystyle AX = AB + BC$ . By considering the power of point $\displaystyle A$ with respect to $\displaystyle \omega$ , we see

$\displaystyle AB^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$ . ∎

Introductory

Example Problems

2006 AIME I Problem 1

Common Pythagorean Triples

A Pythagorean Triple is a set of 3 positive integers such that $a^{2}+b^{2}=c^{2}$ , 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:

3-4-5

5-12-13

7-24-25

8-15-17

9-40-41

@@ Line 6: / Line 6: @@
 This is generalized by the [[Geometric inequality#Pythagorean_Inequality | Pythagorean Inequality]] and the [[Law of Cosines]].
+== Proofs ==
+In these proofs, we will let <math> \displaystyle ABC </math> be any right triangle with a right angle at <math> \displaystyle {} C </math>.
+=== Proof 1 ===
+We use <math> \displaystyle [ABC] </math> to denote the area of triangle <math> \displaystyle ABC </math>.
+Let <math> \displaystyle H </math> be the perpendicular to side <math> \displaystyle AB </math> from <math> \displaystyle {} C </math>.
+<center>[[Image:Pyth1.png]]</center>
+Since <math> \displaystyle ABC, CBF, ACF </math> are similar right triangles, and the areas of similar triangles are proportionate to the squares of corresponding side lengths,
+<center>
+<math> \frac{[ABC]}{AB^2} = \frac{[CBF]}{CB^2} = \frac{[ACF]}{AC^2} </math>.
+</center>
+But since triangle <math> \displaystyle ABC </math> is composed of triangles <math> \displaystyle CBF </math> and <math> \displaystyle ACF </math>, <math> \displaystyle [ABC] = [CBF] + [ACF] </math>, so <math> \displaystyle AB^2 = CB^2 + AC^2 </math>.  {{Halmos}}
+=== Proof 2 ===
+Consider a circle <math> \displaystyle \omega </math> with center <math> \displaystyle B </math> and radius <math> \displaystyle BC </math>.  Since <math> \displaystyle BC </math> and <math> \displaystyle AC </math> are perpendicular, <math> \displaystyle AC </math> is tangent to <math> \displaystyle \omega </math>.  Let the line <math> \displaystyle AB </math> meet <math> \displaystyle \omega </math> at <math> \displaystyle Y </math> and <math> \displaystyle X </math>, as shown in the diagram:
+<center>[[Image:Pyth2.png]]</center>
+Evidently, <math> \displaystyle AY = AB - BC </math> and <math> \displaystyle AX = AB + BC </math>.  By considering the [[power of a point | power]] of point <math> \displaystyle A </math> with respect to <math> \displaystyle \omega </math>, we see
+<center>
+<math> \displaystyle AB^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 </math>.  {{Halmos}}
+</center>
 == Introductory ==
 === Example Problems ===
 * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]

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