Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Pythagorean Theorem"

m (→‎Proof 2)
(took off stub message)
Line 1: Line 1:
−
{{stub}}
 
−
 
  
 
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 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 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 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.  
Line 8: Line 6:
 
== Proofs ==
 
== Proofs ==
  
−
In these proofs, we will let <math> \displaystyle ABC </math> be any right triangle with a right angle at <math> \displaystyle {} C </math>.
+
In these proofs, we will let <math>ABC </math> be any right triangle with a right angle at <math>{} C </math>.
  
 
=== Proof 1 ===
 
=== Proof 1 ===
  
−
We use <math> \displaystyle [ABC] </math> to denote the area of triangle <math> \displaystyle ABC </math>.
+
We use <math>[ABC] </math> to denote the area of triangle <math>ABC </math>.
  
−
Let <math> \displaystyle H </math> be the perpendicular to side <math> \displaystyle AB </math> from <math> \displaystyle {} C </math>.
+
Let <math>H </math> be the perpendicular to side <math>AB </math> from <math>{} C </math>.
  
 
<center>[[Image:Pyth1.png]]</center>
 
<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,
+
Since <math>ABC, CBF, ACF </math> are similar right triangles, and the areas of similar triangles are proportionate to the squares of corresponding side lengths,
 
<center>
 
<center>
 
<math> \frac{[ABC]}{AB^2} = \frac{[CBF]}{CB^2} = \frac{[ACF]}{AC^2} </math>.
 
<math> \frac{[ABC]}{AB^2} = \frac{[CBF]}{CB^2} = \frac{[ACF]}{AC^2} </math>.
 
</center>
 
</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}}
+
But since triangle <math>ABC </math> is composed of triangles <math>CBF </math> and <math>ACF </math>, <math>[ABC] = [CBF] + [ACF] </math>, so <math>AB^2 = CB^2 + AC^2 </math>.  {{Halmos}}
  
 
=== Proof 2 ===
 
=== 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:
+
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>
 
<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
+
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> \displaystyle 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>
  

Revision as of 21:35, 22 September 2007

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

Proof 1

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

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

Pyth1.png

Since $ABC, CBF, ACF$ 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 $ABC$ is composed of triangles $CBF$ and $ACF$, $[ABC] = [CBF] + [ACF]$, so $AB^2 = CB^2 + AC^2$. ∎

Proof 2

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 point $A$ with respect to $\omega$, we see

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

Introductory

Example Problems

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Pythagorean_Theorem&oldid=16551"