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Next: 5. Area Up: 4. Analytic Geometry Previous: 4.2 Reflections, Rotations and   Index

4.3 The Pythagorean Theorem and Distance.

Even though you are probably familiar with the Pythagorean theorem, the result is so important and non-obvious that I am including a proof of it.

4.22   Theorem (Pythagorean Theorem.) In any right triangle, the square on the hypotenuse is equal to the sum of the squares on the two legs.

Proof: Consider a right triangle $T$ whose legs have length $b$ and $c$, and whose hypotenuse has length $a$, and whose angles are $\phi$ and $\theta$ as shown in the figure.


We have $\phi +\theta =90^\circ$ since $T$ is a right triangle.

Construct a square $ABCD$ with sides of length $b+c$, and find points $P,Q,R,S$ dividing the sides of $ABCD$ into pieces of sizes $b$ and $c$ as shown in figure 1. Draw the lines $PQ,QR,RS$, and $SP$, thus creating four triangles congruent to $T$ (i.e., four right triangles with legs of length $b$ and $c$). Each angle of $PQRS$ is $180^\circ -(\phi +\theta)=180^\circ -90^\circ =90^\circ$ so $PQRS$ is a square of side $a$. The four triangles in figure 1 each have area $\displaystyle {{1\over 2}bc}$, so
\mbox{\rm area}(ABCD)-4 \cdot\mbox{\rm area}(T)=a^2
\end{displaymath} (4.23)


\begin{displaymath}(b+c)^2 -2bc = a^2 \end{displaymath}

and hence
b^2 + c^2 =a^2
\mbox{ $\diamondsuit$}
\end{displaymath} (4.24)

The proof just given uses a combination of algebra and geometry. I will now give a second proof that is completely geometrical.

Construct a second square $WXYZ$ with sides of length $b+c$, and mark off segments $WE$ and $WF$ of length $c$ as shown in figure 2. Then draw $EK$ perpendicular to $WX$ and let $EK$ intersect $ZY$ at $G$, and draw $FL$ perpendicular to $WZ$ and let $FL$ intersect $XY$ at $H$. Then $EGZ$ is a right angle, since the other angles of the quadrilateral $WEGZ$ are right angles. Similarly angle $FHX$ is a right angle. Thus $WEGZ$ is a rectangle so $ZG=c$ and similarly $WFHX$ is a rectangle and $XH=c$. Moreover $EG$ and $FH$ are perpendicular since $EG\Vert WZ$ and $FH\Vert WX$. Thus the region labeled $S_1$ is a square with side $c$ and the region labeled $S_2$ is a square with side $b$.

In figure 2 we have $\mbox{\rm area}(R_1)=\mbox{\rm area}(R_2)=2 \mbox{\rm area}(T)$, and hence

\mbox{\rm area}(WXYZ) - 4\cdot\mbox{\rm area}(T) = b^2 + c^2.
\end{displaymath} (4.25)

We have $\mbox{\rm area}(ABCD) = \mbox{\rm area}(WXYZ)$ since $ABCD$ and $WXYZ$ are both squares with side $b+c$. Hence from equations (4.23) and (4.25) we see that

\begin{displaymath}a^2 = b^2 + c^2.\mbox{ $\diamondsuit$}\end{displaymath}

Although the theorem we just proved is named for Pythagoras (fl. 530-510 B.C) , it was probably known much earlier. There is evidence that it was known to the Babylonians circa 1000 BC[27, pp 118-121]. Legend has it that

Emperor Yu[circa 21st century B.C.] quells floods, he deepens rivers and streams, observes the shape of mountains and valleys, surveys the high and low places, relieves the greatest calamities and saves the people from danger. He leads the floods east into the sea and ensures no flooding or drowning. This is made possible because of the Gougu theorem $\ldots$[47, page 29].

\put(2.7,1){G\={o}ug\v{u} shape}

``Gougu'' is the shape shown in the figure, and the Gougu theorem is our Pythagorean theorem. The prose style here is similar to that of current day mathematicians trying to get congress to allocate funds for the support of mathematics.

Katyayana(c. 600 BC or 500BC??) stated the general theorem:

The rope [stretched along the length] of the diagonal of a rectangle makes an [area] which the vertical and horizontal sides make together.[27, page 229]

4.26   Theorem (Distance formula.) If $\mbox{{\bf a}}$ and $\mbox{{\bf b}}$ are points in $\mbox{{\bf R}}^2$ then the distance from $\mbox{{\bf a}}$ to $\mbox{{\bf b}}$ is

\begin{displaymath}d(\mbox{{\bf a}},\mbox{{\bf b}})=\sqrt{(a_1-b_1)^2+(a_2-b_2)^2}.\end{displaymath}

Proof: Draw the vertical line through $\mbox{{\bf a}}$ and the horizontal line through $\mbox{{\bf b}}$. These lines intersect at the point $\mbox{{\bf p}}= (a_1,b_2)$. The length of $[\mbox{{\bf a}}\mbox{{\bf p}}]$ is $\vert a_2-b_2\vert$ and the length of $[\mbox{{\bf p}}\mbox{{\bf b}}]$ is $\vert a_1-b_1\vert$ and $[\mbox{{\bf a}}\mbox{{\bf b}}]$ is the hypotenuse of a right angle with legs $[\mbox{{\bf a}}\mbox{{\bf p}}]$ and $[\mbox{{\bf p}}\mbox{{\bf b}}]$.


By the Pythagorean theorem,

\begin{displaymath}\left(\mbox{length}([\mbox{{\bf a}}\mbox{{\bf b}}])\right)^2=(a_1-b_1)^2 +(a_2-b_2)^2\end{displaymath}

so length $([\mbox{{\bf a}},\mbox{{\bf b}}])=\sqrt{(a_1-b_1)^2 +(a_2-b_2)^2}$. $\diamondsuit$

4.27   Notation ( $d(\mbox{{\bf a}},\mbox{{\bf b}})$, distance $(\mbox{{\bf a}},\mbox{{\bf b}})$) If $\mbox{{\bf a}}$ and $\mbox{{\bf b}}$ are points in $\mbox{{\bf R}}^2$, I will denote the distance from $\mbox{{\bf a}}$ to $\mbox{{\bf b}}$ by either distance $(\mbox{{\bf a}},\mbox{{\bf b}})$ or by $d(\mbox{{\bf a}},\mbox{{\bf b}})$.

4.28   Definition (Circle.) Let ${\bf p}=(a,b)$ be a point in $\mbox{{\bf R}}^2$, and let $r\in\mbox{${\mbox{{\bf R}}}^{+}$}$. The circle with center ${\bf p}$ and radius $r$ is defined to be

C({\bf p},r)&=&\{(x,y)\in\mbox{{\bf R}}^2\colon d((x,y),(a,b))...
...} \\
&=&\{(x,y)\in\mbox{{\bf R}}^2\colon (x-a)^2+(y-b)^2=r^2\}.

The equation


is called the equation of the circle $C({\bf p},r)$. The circle $C((0,0),1)$ is called the unit circle.

We will now review the method for solving quadratic equations.

4.29   Theorem (Quadratic formula.) Let $A$, $B$, and $C$ be real numbers with $A \neq 0$.

If $B^2 - 4AC <0$, then the equation $Ax^2 + Bx + C = 0$ has no solutions in R.

If $B^2 - 4AC \geq 0$, then the set of solutions of the equation $Ax^2 + Bx + C = 0$ is

\left\{ {-B \pm \sqrt{B^2-4AC} \over 2A}\right\}.
\end{displaymath} (4.30)

The set (4.30) contains one or two elements, depending on whether $B^2 - 4AC$ is zero or positive.)

Proof: Let $A,B,C$ be real numbers with $A \neq 0$. Let $x \in \mbox{{\bf R}}$. Then

Ax^2+Bx+C=0 &\iff& A\Big(x^2+{{Bx}\over A}\Big)=-C \\
&\iff& ...
...} \\
&\iff& \Big(x+{B\over {2A}}\Big)^2={{B^2-4AC}\over {4A^2}}

Hence $Ax^2 + Bx + C = 0$ has no solutions unless $B^2 - 4AC \geq 0$. If $B^2 - 4AC \geq 0$, then the solutions are given by

\begin{displaymath}\Big( x+{B\over {2A}}\Big) = {{ {\pm}\sqrt{B^2-4AC}}\over {2A}}\end{displaymath}


\begin{displaymath}x={{-B\pm \sqrt{B^2-4AC}}\over {2A}}.\mbox{ $\diamondsuit$}\end{displaymath}

4.31   Example. Describe the set $C((0,0),6)\cap C((4,4),2)$.
The sketch suggests that this set will consist of two points in the first quadrant. Let $(x,y)$ be a point in the intersection. Then
\end{displaymath} (4.32)

(x-4)^2+(y-4)^2=4,\; {\rm i.e.}\; x^2+y^2-8x-8y+28=0.
\end{displaymath} (4.33)

It follows that $36-8x-8y+28=0$, or $8x+8y-64=0$ or
\end{displaymath} (4.34)

(The line whose equation is $y=8-x$ is shown in the figure. We've proved that the intersection is a subset of this line.) Replace $y$ by $8-x$ in equation (4.33) to obtain








By the quadratic formula, it follows that

\begin{displaymath}x={{8\pm\sqrt{64-56}}\over 2}={{8\pm 2\sqrt 2}\over 2}=4\pm\sqrt 2.\end{displaymath}

By equation (4.34)

\begin{displaymath}y=8-x=4\mp\sqrt 2.\end{displaymath}

We have shown that if $(x,y)\in C((0,0),6)\cap C((4,4),2)$, then
$(x,y)\in\{(4+\sqrt 2,4-\sqrt 2), (4-\sqrt 2,4+\sqrt 2)\}$. It is easy to verify that each of the two calculated points satisfies both equations  (4.32) and (4.33) so

\begin{displaymath}C((0,0),6)\cap C((4,4),2)=\{(4+\sqrt 2,4-\sqrt 2),(4-\sqrt 2,4+\sqrt

next up previous index
Next: 5. Area Up: 4. Analytic Geometry Previous: 4.2 Reflections, Rotations and   Index
Ray Mayer 2007-09-07