Pythagorean Theorem

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:

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

Consider a square of side length $a + b$ . Inside it, place four congruent right triangles—each with legs $a$ and $b$ and hypotenuse $c$ —arranged so that their right angles occupy the corners of the large square and their hypotenuses form a smaller, tilted square at the center.

$[asy] unitsize(3cm); // define a and b real a = 0.6; real b = 0.8; // corners of the big square pair A = (0,0); pair B = (a+b, 0); pair C = (a+b, a+b); pair D = (0, a+b); draw(A--B--C--D--cycle); // four right triangles pair T1 = (a,0); pair T2 = (a+b, b); pair T3 = (b, a+b); pair T4 = (0, a); draw(A--T1--T4--cycle); // bottom-left triangle draw(T1--B--T2--cycle); // bottom-right triangle draw(T4--T3--D--cycle); // top-left triangle draw(T2--C--T3--cycle); // top-right triangle // inner square formed by hypotenuses draw(T1--T2--T3--T4--cycle); // labels label("$a+b$", midpoint(A--B), S); label("$a+b$", midpoint(B--C), E); label("$a$", midpoint(A--T1), S); label("$b$", midpoint(T1--B), S); label("$b$", midpoint(D--T4), W); label("$a$", midpoint(T4--A), W); label("$c$", midpoint(T1--T2), NE); label("$c$", midpoint(T2--T3), NE); label("$c$", midpoint(T3--T4), NW); label("$c$", midpoint(T4--T1), NW); [/asy]$

The area of the large square is $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ . The area can also be expressed as the sum of the four triangles, each with area $\frac{1}{2} a b$ , totaling $2 a b$ , plus the inner square with area $c^{2}$ . Setting these equal gives: $(a + b)^2 = 4 \times \frac{1}{2} ab + c^2 = 2ab + c^2.$ Simplifying, $a^2 + 2ab + b^2 = 2ab + c^2 \implies a^2 + b^2 = c^2. \qquad\blacksquare$

Proof 5 (Linear Algebra / Inner Product Space)

Consider the right triangle $\triangle ABC$ with a right angle at $C$ . Represent the vectors $\vec{AC} = \mathbf{u} \quad \text{and} \quad \vec{BC} = \mathbf{v}$ in an inner product space equipped with the dot product $\langle \cdot, \cdot \rangle$ .

Because $\angle ACB$ is a right angle, the vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal: $\langle \mathbf{u}, \mathbf{v} \rangle = 0.$

By definition of the norm induced by the inner product, $\|\mathbf{u}\| = a, \quad \|\mathbf{v}\| = b,$ and the hypotenuse vector is $\vec{AB} = \mathbf{u} - \mathbf{v}.$

Calculate the squared length of $\vec{AB}$ : $\|\vec{AB}\|^2 = \langle \mathbf{u} - \mathbf{v}, \mathbf{u} - \mathbf{v} \rangle = \langle \mathbf{u}, \mathbf{u} \rangle - 2 \langle \mathbf{u}, \mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{v} \rangle.$

Using orthogonality $\langle \mathbf{u}, \mathbf{v} \rangle = 0$ , this reduces to $\|\vec{AB}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 = a^2 + b^2.$

Since $\|\vec{AB}\| = c$ , the length of the hypotenuse, we conclude $c^2 = a^2 + b^2,$ which is the Pythagorean Theorem.

This proof is highly abstract and depends on familiarity with vector spaces and inner products, moving far beyond the classical Euclidean framework into the realm of modern algebraic geometry.

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 $\frac{n-1}{2} + 1$ .

Problems

Introductory

Problem 1

Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$

$[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); [/asy]$

$\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$

Solution

Problem 2

A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?

$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$