Difference between revisions of "2013 AIME II Problems/Problem 8"

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Problem 8

A hexagon that is inscribed in a circle has side lengths $22$ , $22$ , $20$ , $22$ , $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ .

Solution

$[asy] import olympiad; import math; pair A,B,C,D,E,F; B=origin; C=(10,0); D=(12,-5); E=(10,-10); F=(0,-10); A=(-2, -5); draw(A--B);draw(B--C);draw(C--D);draw(D--E);draw(E--F);draw(F--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(circumcircle(A, D, F)); label("$A$",A,W);label("$B$",B,W);label("$C$",C,N);label("$D$",D,W);label("$E$",E,S);label("$F$",F,W); [/asy]$

Solution 1

Let us call the hexagon $ABCDEF$ , where $AB=CD=DE=AF=22$ , and $BC=EF=20$ . We can just consider one half of the hexagon, $ABCD$ , to make matters simpler. Draw a line from the center of the circle, $O$ , to the midpoint of $BC$ , $E$ . Now, draw a line from $O$ to the midpoint of $AB$ , $F$ . Clearly, $\angle BEO=90^{\circ}$ , because $BO=CO$ , and $\angle BFO=90^{\circ}$ , for similar reasons. Also notice that $\angle AOE=90^{\circ}$ . Let us call $\angle BOF=\theta$ . Therefore, $\angle AOB=2\theta$ , and so $\angle BOE=90-2\theta$ . Let us label the radius of the circle $r$ . This means $\sin{\theta}=\frac{BF}{r}=\frac{11}{r}$ $\sin{(90-2\theta)}=\frac{BE}{r}=\frac{10}{r}$ Now we can use simple trigonometry to solve for $r$ . Recall that $\sin{(90-\alpha)}=\cos(\alpha)$ : That means $\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}$ . Recall that $\cos{2\alpha}=1-2\sin^2{\alpha}$ : That means $\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}$ . Let $\sin{\theta}=x$ . Substitute to get $x=\frac{11}{r}$ and $1-2x^2=\frac{10}{r}$ Now substitute the first equation into the second equation: $1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}$ Multiplying both sides by $r^2$ and reordering gives us the quadratic $r^2-10r-242=0$ Using the quadratic equation to solve, we get that $r=5+\sqrt{267}$ (because $5-\sqrt{267}$ gives a negative value), so the answer is $5+267=\boxed{272}$ .

Solution 2

Using the trapezoid $ABCD$ mentioned above, draw an altitude of the trapezoid passing through point $B$ onto $AD$ at point $E$ . Now, we can use the pythagorean theorem: $(22^2-(r-10)^2)+10^2=r^2$ . Expanding and combining like terms gives us the quadratic $r^2-10r-242=0$ and solving for $r$ gives $r=5+\sqrt{267}$ . So the solution is $5+267=\boxed{272}$ .

Solution 3

Join the diameter of the circle $AD$ and let the length be $d$ . By Ptolemy's Theorem on trapezoid $ADEF$ , $(AD)(EF) + (AF)(DE) = (AE)(DF)$ . Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to $x$ each. Then

$20d + 22^2 = x^2$

Since $\angle AED$ is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right $\triangle AED$ :

$(AE)^2 + (ED)^2 = (AD)^2$ $x^2 + 22^2 = d^2$

From the above equations, we have: $x^2 = d^2 - 22^2 = 20d + 22^2$ $d^2 - 20d = 2\times22^2$ $d^2 - 20d + 100 = 968+100 = 1068$ $(d-10) = \sqrt{1068}$ $d = \sqrt{1068} + 10 = 2\times(\sqrt{267}+5)$

Since the radius is half the diameter, it is $\sqrt{267}+5$ , so the answer is $5+267 \Rightarrow \boxed{272}$ .

2013 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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