Difference between revisions of "Pythagorean Theorem"

Latest revision as of 09:44, 17 January 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:

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 triple 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, $6-8-10 = (3-4-5)*2$ $21-72-75 = (7-24-25)*3$ $10-24-26 = (5-12-13)*2$

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 $11$ , 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 $4+5=\boxed{9}$ .

External links

122 proofs of the Pythagorean Theorem

@@ Line 1: / Line 1: @@
-The '''Pythagorean Theorem''' states that for a [[right triangle]] with legs of length <math>a</math> and <math>b</math> and [[hypotenuse]] of length <math>c</math> we have the relationship <math>{a}^{2}+{b}^{2}={c}^{2}</math>.  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.
+The '''Pythagorean Theorem''' states that for a [[right triangle]] with [[leg]]s of length <math>a</math> and <math>b</math> and [[hypotenuse]] of length <math>c</math> we have the relationship <math>a^2+b^2=c^2</math>.  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 [[Geometric inequality#Pythagorean_Inequality | Pythagorean Inequality]] and the [[Law of Cosines]].
+This is generalized by the [[Geometric inequality#Pythagorean_Inequality|Pythagorean Inequality]] and the [[Law of Cosines]].
 == Proofs ==
-In these proofs, we will let <math>ABC </math> be any right triangle with a right angle at <math>\angle ACB</math>.
+In these proofs, we will let <math>ABC</math> be any right triangle with a right angle at <math>\angle ACB</math>, and we use <math>[ABC]</math> to denote the area of triangle <math>ABC</math>.
 === Proof 1 ===
-We use <math>[ABC] </math> to denote the area of triangle <math>ABC </math>.
+Let <math>D</math> be the foot of the [[altitude]] from <math>C</math>. <math>ABC</math>, <math>ACD</math>, <math>BCD</math> are similar triangles, so <math>\frac{AC}{AD}=\frac{AB}{AC} \implies AC^2=(AD)(AB)</math> and <math>\frac{BC}{BD}=\frac{AB}{BC} \implies BC^2=(BD)(AB)</math>. Adding these equations gives us <math>AC^2+BC^2=(AD)(AB)+(BD)(AB) \implies AC^2+BC^2=(AB)(AD+BD) \implies AC^2+BC^2=AB^2</math>
-Let <math>H </math> be the perpendicular to side <math>AB </math> from <math>{} C </math>.
+=== Proof 2 ===
+Let <math>H</math> be the foot of the [[altitude]] from <math>C</math>.
 <center>
@@ Line 32: / Line 34: @@
 </center>
-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,
+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,
-<center>
+<cmath>\frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2}.</cmath>
-<math> \frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2} </math>.
+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}}
-</center>
-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}}
-=== Proof 2 ===
+=== Proof 3 ===
-Consider a circle <math>\omega </math> with center <math>B </math> and radius <math>BC </math>.  Since <math>BC </math> and <math>AC </math> are perpendicular, <math>AC </math> is tangent to <math>\omega </math>.  Let the line <math>AB </math> meet <math>\omega </math> at <math>Y </math> and <math>X </math>, as shown in the diagram:
+Consider a circle <math>\omega</math> with center <math>B</math> and radius <math>BC</math>.  Since <math>BC</math> and <math>AC</math> are perpendicular, <math>AC</math> is tangent to <math>\omega</math>.  Let the line <math>AB</math> meet <math>\omega</math> at <math>Y</math> and <math>X</math>, as shown in the diagram:
 <center>[[Image:Pyth2.png]]</center>
-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
+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}}
-<center>
-<math>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 </math>.  {{Halmos}}
-</center>
-=== Proof 3 ===
+=== Proof 4 ===
 <math>ABCD</math> and <math>EFGH</math> are squares.
 <center>
 <asy>
-pair A, B,C,D;
+pair A,B,C,D;
 A = (-10,10);
 B = (10,10);
@@ Line 93: / Line 90: @@
 </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}}
+<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}}
+== Pythagorean Triples ==
-== Common Pythagorean Triples ==
+{{main|Pythagorean triple}}
-A [[Pythagorean Triple]] is a [[set]] of 3 [[positive integer]]s such that <math>a^{2}+b^{2}=c^{2}</math>, i.e. the 3 numbers can be the lengths of the sides of a right triangle.  Among these, the [[Primitive Pythagorean Triple]]s, those in which the three numbers have no common [[divisor]], are most interesting.  A few of them are:
+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:
 <cmath>3-4-5</cmath>
@@ Line 107: / Line 106: @@
 <cmath>11-60-61</cmath>
+Note that (3,4,5) is the only Pythagorean triple that consists of consecutive integers.
-Also Pythagorean Triples can be created with the a Pythagorean triple by multiplying the lengths by any integer.
+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,
 <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>
-Note that (-1,0,1) and (3,4,5) are the only pythagoren triplets that consist of consecutive integers.
-Also, if (a,b,c) are a pythagorean triplet it follows that (ka,kb,kc) will also form a pythagorean triplet for any constant k.
+== Problems ==
-k can also be imaginary.
-== Problems ==
 === Introductory ===
 * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]
 * [[2007 AMC 12A Problems/Problem 10 | 2007 AMC 12A Problem 10]]
-=== Sample Problem ===
+=== Problem 1 ===
 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>.
+<math>\sqrt{333^2+444^2} = \boxed{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>.
+<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>.
-=== Another Problem ===
+=== Problem 2 ===
 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) ====
@@ Line 142: / Line 138: @@
 is a leg and 4 is the hypotenuse.
-There are no more cases as the hypotenuse has to be greater than the leg.
+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>.
-This makes the sum <math>4+5=9</math>.
 == External links ==
@@ Line 150: / Line 144: @@
 [[Category:Geometry]]
 [[Category:Theorems]]

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