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

(→‎Solution 2)
(→‎Solution 3: Add solution 3 using Ptolemy's theorem)
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===Solution 2===
 
===Solution 2===
 
Using the trapezoid <math>ABCD</math> mentioned above, draw an altitude of the trapezoid passing through point <math>B</math> onto <math>AD</math> at point <math>E</math>. Now, we can use the pythagorean theorem: <math>(22^2-(r-10)^2)+10^2=r^2</math>. Expanding and combining like terms gives us the quadratic <cmath>r^2-10r-242=0</cmath> and solving for <math>r</math> gives <math>r=5+\sqrt{267}</math>. So the solution is <math>5+267=\boxed{272}</math>
 
Using the trapezoid <math>ABCD</math> mentioned above, draw an altitude of the trapezoid passing through point <math>B</math> onto <math>AD</math> at point <math>E</math>. Now, we can use the pythagorean theorem: <math>(22^2-(r-10)^2)+10^2=r^2</math>. Expanding and combining like terms gives us the quadratic <cmath>r^2-10r-242=0</cmath> and solving for <math>r</math> gives <math>r=5+\sqrt{267}</math>. So the solution is <math>5+267=\boxed{272}</math>
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 +
===Solution 3===
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 +
Join the diameter of the circle <math>AD</math> and let the length be <math>d</math>. By [[Ptolemy's Theorem]] on trapezoid <math>ADEF</math>, <math>(AD)(EF) + (AF)(DE) = (AE)(DF)</math>. Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to <math>x</math> each. Then
 +
 +
<cmath>20d + 22^2 = x^2</cmath>
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Since <math>\angle AED</math> is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right <math>\triangle AED</math>:
 +
 +
<cmath>(AE)^2 + (ED)^2 = (AD)^2</cmath>
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<cmath>x^2 + 22^2 = d^2</cmath>
 +
 +
From the above equations, we have:
 +
<cmath>x^2 = d^2 - 22^2 = 20d + 22^2</cmath>
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<cmath>d^2 - 20d = 2\times22^2</cmath>
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<cmath>d^2 - 20d + 100 = 968+100 = 1068</cmath>
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<cmath>(d-10) = \sqrt{1068}</cmath>
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<cmath>d = \sqrt{1068} + 10 = 2\times(\sqrt{267}+5)</cmath>
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 +
Since the radius is half the diameter, it is <math>\sqrt{267}+5</math>, so the answer is <math>5+267 \Rightarrow \boxed{272}</math>.

Revision as of 22:13, 4 April 2013

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

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 $\angle BFO=\theta$. Therefore, $\angle AOB=2\theta$, and so $\angle AOE=90-2\theta$. 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 $5+267 \Rightarrow \boxed{272}$.

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