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 "2013 AIME II Problems/Problem 8"

(→‎Solution)
(→‎Solution 5)
 
(36 intermediate revisions by 15 users not shown)
Line 1: Line 1:
 +
==Problem 8==
 
A hexagon that is inscribed in a circle has side lengths <math>22</math>, <math>22</math>, <math>20</math>, <math>22</math>, <math>22</math>, and <math>20</math> in that order. The radius of the circle can be written as <math>p+\sqrt{q}</math>, where <math>p</math> and <math>q</math> are positive integers. Find <math>p+q</math>.
 
A hexagon that is inscribed in a circle has side lengths <math>22</math>, <math>22</math>, <math>20</math>, <math>22</math>, <math>22</math>, and <math>20</math> in that order. The radius of the circle can be written as <math>p+\sqrt{q}</math>, where <math>p</math> and <math>q</math> are positive integers. Find <math>p+q</math>.
  
 
==Solution==
 
==Solution==
 +
<asy>
 +
import olympiad;
 +
import math;
  
 +
real a;
 +
a=2*asin(11/(5+sqrt(267)));
 +
pair A,B,C,D,E,F;
 +
A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a);
 +
draw(A--B--C--D--E--F--A--D);
 +
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F);
 +
draw(unitcircle);
 +
label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D,dir(0));label("$E$",E,SE);label("$F$",F,SW);
 +
 +
</asy>
 +
 +
 +
===Solution 1===
 
Let us call the hexagon <math>ABCDEF</math>, where <math>AB=CD=DE=AF=22</math>, and <math>BC=EF=20</math>.  
 
Let us call the hexagon <math>ABCDEF</math>, where <math>AB=CD=DE=AF=22</math>, and <math>BC=EF=20</math>.  
 
We can just consider one half of the hexagon, <math>ABCD</math>, to make matters simpler.  
 
We can just consider one half of the hexagon, <math>ABCD</math>, to make matters simpler.  
−
Draw a line from the center of the circle, <math>O</math>, to the midpoint of <math>BC</math>, <math>E</math>. Now, draw a line from <math>O</math> to the midpoint of <math>AB</math>, <math>F</math>. Clearly, <math>\angle BEO=90^{\circ}</math>, because <math>BO=CO</math>, and <math>\angle BFO=90^{\circ}</math>, for similar reasons. Also notice that <math>\angle AOE=90^{\circ}</math>.  
+
Draw a line from the center of the circle, <math>O</math>, to the midpoint of <math>BC</math>, <math>X</math>. Now, draw a line from <math>O</math> to the midpoint of <math>AB</math>, <math>Y</math>. Clearly, <math>\angle BXO=90^{\circ}</math>, because <math>BO=CO</math>, and <math>\angle BYO=90^{\circ}</math>, for similar reasons. Also notice that <math>\angle AOX=90^{\circ}</math>.  
−
Let us call <math>\angle BFO=\theta</math>. Therefore, <math>\angle AOB=2\theta</math>, and so <math>\angle AOE=90-2\theta</math>. Let us label the radius of the circle <math>r</math>. This means <cmath>\sin{\theta}=\frac{BF}{r}=\frac{11}{r}</cmath> <cmath>\sin{90-2\theta}=\frac{BE}{r}=\frac{10}{r}</cmath>
+
Let us call <math>\angle BOY=\theta</math>. Therefore, <math>\angle AOB=2\theta</math>, and so <math>\angle BOX=90-2\theta</math>. Let us label the radius of the circle <math>r</math>. This means <cmath>\sin{\theta}=\frac{BY}{r}=\frac{11}{r}</cmath> <cmath>\sin{(90-2\theta)}=\frac{BX}{r}=\frac{10}{r}</cmath>
 
Now we can use simple trigonometry to solve for <math>r</math>.
 
Now we can use simple trigonometry to solve for <math>r</math>.
−
Recall that  <math>\sin{90-\alpha}=\cos(\alpha)</math>: That means <math>\sin{90-2\theta}=\cos{2\theta}=\frac{10}{r}</math>
+
Recall that  <math>\sin{(90-\alpha)}=\cos(\alpha)</math>: That means <math>\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}</math>.
 
Recall that <math>\cos{2\alpha}=1-2\sin^2{\alpha}</math>: That means <math>\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}</math>.
 
Recall that <math>\cos{2\alpha}=1-2\sin^2{\alpha}</math>: That means <math>\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}</math>.
 
Let <math>\sin{\theta}=x</math>.
 
Let <math>\sin{\theta}=x</math>.
Line 14: Line 31:
 
Now substitute the first equation into the second equation: <math>1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}</math>
 
Now substitute the first equation into the second equation: <math>1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}</math>
 
Multiplying both sides by <math>r^2</math> and reordering gives us the quadratic <cmath>r^2-10r-242=0</cmath>
 
Multiplying both sides by <math>r^2</math> and reordering gives us the quadratic <cmath>r^2-10r-242=0</cmath>
−
Using the quadratic equation to solve, we get that <math>r=5+\sqrt{267}</math>, so the answer is <math>5+267=\boxed{272}</math>
+
Using the quadratic equation to solve, we get that <math>r=5+\sqrt{267}</math> (because <math>5-\sqrt{267}</math> gives a negative value), so the answer is <math>5+267=\boxed{272}</math>.
 +
 
 +
===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>J</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>.
 +
 
 +
===Solution 3===
 +
 
 +
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>
 +
 
 +
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>
 +
<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>
 +
<cmath>d^2 - 20d = 2\times22^2</cmath>
 +
<cmath>d^2 - 20d + 100 = 968+100 = 1068</cmath>
 +
<cmath>(d-10) = \sqrt{1068}</cmath>
 +
<cmath>d = \sqrt{1068} + 10 = 2\times(\sqrt{267}+5)</cmath>
 +
 
 +
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>.
 +
===Solution 4===
 +
As we can see this image, it is symmetrical hence the diameter divides the hexagon into two congruent quadrilateral. Now we can apply the Ptolemy's theorem. Denote the radius is r, we can get <cmath>22*2x+440=\sqrt{4x^2-400}\sqrt{4x^2-484}</cmath>, after simple factorization, we can get <cmath>x^4-342x^2-2420x=0</cmath>, it is easy to see that <math>x=-10, x=0</math> are two solutions for the equation, so we can factorize that into <cmath>x(x+10)(x^2-10x-242)</cmath>so we only need to find the solution for <cmath>x^2-10x-242=0</cmath> and we can get <math>x=(\sqrt{267}+5)</math> is the desired answer for the problem, and our answer is <math>5+267 \Rightarrow \boxed{272}</math>.～bluesoul
 +
 
 +
===Solution 6 (Trig Bash)===
 +
 
 +
Let <math>\angle{AOB} = \theta</math>. So, we have <math>\sin \dfrac{\theta}{2} = \dfrac{11}{r}</math> and <math>\cos \dfrac{\theta}{2} = \dfrac{\sqrt{r^{2} - 121}}{r}</math>. So, <math>\sin \theta = 2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} = \dfrac{22 \sqrt{r^{2} - 121}}{r^{2}}</math>. Let <math>H</math> be the foot of the perpendicular from <math>B</math> to <math>\overline{AD}</math>. We have <math>BF = 2 BH = 2 r \sin \theta = \dfrac{44 \sqrt{r^{2} - 121}}{r}</math>. Using Pythagorean theorem on <math>\triangle BCF</math>, to get <math>(\dfrac{44 \sqrt{r^{2} - 121}}{r})^{2} + 20^{2} = (2r)^{2}</math>, or <math>\dfrac{44^{2}r^{2} - 44^{2} \cdot 121}{r^{2}} + 20^{2} = 4r^{4}</math>. Multiplying by <math>r^{2}</math>, we get <math>44^{2} r^{2} - 44^{2} \cdot 121 + 20^{2} r^{2} = 4r^{4}</math>. Rearranging and simplifying, we get a quadratic in <math>r^{2}</math>: <cmath>r^{4} - 584r^{2} + 242^{2} = 0 \text{,}</cmath> which gives us <math>r^{2} = 292 \pm 10\sqrt{267}</math>. Because <math>r</math> is in the form <math>p + \sqrt{q}</math>, we know to choose the larger option, meaning <math>r^2 = 292 + 10\sqrt{267}</math>, so <math>p\sqrt{q} = 5\sqrt{267}</math> and <math>p^2 + q = 292</math>. By inspection, we get <math>(p, q) = (5, 267)</math>, so our answer is <math>5 + 267 = \boxed{272}</math>.
 +
 
 +
~Puck_0
 +
==Solution 7==
 +
<asy>
 +
import olympiad;
 +
import math;
 +
 
 +
real a;
 +
a=2*asin(11/(5+sqrt(267)));
 +
pair A,B,C,D,E,F;
 +
A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a);
 +
draw(A--B--C--D--E--F--A--D);
 +
draw(B--D);
 +
draw(A--C);
 +
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F);
 +
draw(unitcircle);
 +
label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D,dir(0));label("$E$",E,SE);label("$F$",F,SW);
 +
 
 +
</asy>
 +
 
 +
We know that <math>AD=x</math> is a diameter, hence <math>ABD</math> and <math>ACD</math> are right triangles. Let <math>AB=BC=22</math>, and <math>CD=20.</math> Hence, <math>ABD</math> is a right triangle with legs <math>22,\sqrt{x^2-484},</math> and hypotenuse, <math>x,</math> and <math>ACD</math> is a right triangle with legs <math>20, \sqrt{x^2-400},</math> with hypotenuse <math>x</math>. By Ptolemy's we have <cmath>22(x+20)=\sqrt{x^2-400}\sqrt{x^2-484}</cmath>.
 +
We square both sides to get <cmath>484(x+20)^2=(x^2-400)(x^2-484) \implies 484(x+20)=(x-20)(x^2-484) \implies 484x=x^3-20x^2-484x \implies x(x^2-20x-968)=0</cmath>
 +
 
 +
We solve for <math>x</math> via the Quadratic Formula and receive <math>x=10+2\sqrt{267}</math>, but we must divide by <math>2</math> since we want the radius, and hence <math>267+5=\boxed{272}.</math>
 +
~SirAppel
 +
 
 +
==See Also==
 +
{{AIME box|year=2013|n=II|num-b=7|num-a=9}}
 +
{{MAA Notice}}

Latest revision as of 21:24, 19 July 2024

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; real a; a=2*asin(11/(5+sqrt(267))); pair A,B,C,D,E,F; A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a); draw(A--B--C--D--E--F--A--D); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(unitcircle); label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D,dir(0));label("$E$",E,SE);label("$F$",F,SW); [/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 $BC$, $X$. Now, draw a line from $O$ to the midpoint of $AB$, $Y$. Clearly, $\angle BXO=90^{\circ}$, because $BO=CO$, and $\angle BYO=90^{\circ}$, for similar reasons. Also notice that $\angle AOX=90^{\circ}$. Let us call $\angle BOY=\theta$. Therefore, $\angle AOB=2\theta$, and so $\angle BOX=90-2\theta$. Let us label the radius of the circle $r$. This means \[\sin{\theta}=\frac{BY}{r}=\frac{11}{r}\] \[\sin{(90-2\theta)}=\frac{BX}{r}=\frac{10}{r}\] 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 $J$. 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 $5+267 \Rightarrow \boxed{272}$.

Solution 4

As we can see this image, it is symmetrical hence the diameter divides the hexagon into two congruent quadrilateral. Now we can apply the Ptolemy's theorem. Denote the radius is r, we can get \[22*2x+440=\sqrt{4x^2-400}\sqrt{4x^2-484}\], after simple factorization, we can get \[x^4-342x^2-2420x=0\], it is easy to see that $x=-10, x=0$ are two solutions for the equation, so we can factorize that into \[x(x+10)(x^2-10x-242)\]so we only need to find the solution for \[x^2-10x-242=0\] and we can get $x=(\sqrt{267}+5)$ is the desired answer for the problem, and our answer is $5+267 \Rightarrow \boxed{272}$.～bluesoul

Solution 6 (Trig Bash)

Let $\angle{AOB} = \theta$. So, we have $\sin \dfrac{\theta}{2} = \dfrac{11}{r}$ and $\cos \dfrac{\theta}{2} = \dfrac{\sqrt{r^{2} - 121}}{r}$. So, $\sin \theta = 2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} = \dfrac{22 \sqrt{r^{2} - 121}}{r^{2}}$. Let $H$ be the foot of the perpendicular from $B$ to $\overline{AD}$. We have $BF = 2 BH = 2 r \sin \theta = \dfrac{44 \sqrt{r^{2} - 121}}{r}$. Using Pythagorean theorem on $\triangle BCF$, to get $(\dfrac{44 \sqrt{r^{2} - 121}}{r})^{2} + 20^{2} = (2r)^{2}$, or $\dfrac{44^{2}r^{2} - 44^{2} \cdot 121}{r^{2}} + 20^{2} = 4r^{4}$. Multiplying by $r^{2}$, we get $44^{2} r^{2} - 44^{2} \cdot 121 + 20^{2} r^{2} = 4r^{4}$. Rearranging and simplifying, we get a quadratic in $r^{2}$: \[r^{4} - 584r^{2} + 242^{2} = 0 \text{,}\] which gives us $r^{2} = 292 \pm 10\sqrt{267}$. Because $r$ is in the form $p + \sqrt{q}$, we know to choose the larger option, meaning $r^2 = 292 + 10\sqrt{267}$, so $p\sqrt{q} = 5\sqrt{267}$ and $p^2 + q = 292$. By inspection, we get $(p, q) = (5, 267)$, so our answer is $5 + 267 = \boxed{272}$.

~Puck_0

Solution 7

[asy] import olympiad; import math; real a; a=2*asin(11/(5+sqrt(267))); pair A,B,C,D,E,F; A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a); draw(A--B--C--D--E--F--A--D); draw(B--D); draw(A--C); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(unitcircle); label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D,dir(0));label("$E$",E,SE);label("$F$",F,SW); [/asy]

We know that $AD=x$ is a diameter, hence $ABD$ and $ACD$ are right triangles. Let $AB=BC=22$, and $CD=20.$ Hence, $ABD$ is a right triangle with legs $22,\sqrt{x^2-484},$ and hypotenuse, $x,$ and $ACD$ is a right triangle with legs $20, \sqrt{x^2-400},$ with hypotenuse $x$. By Ptolemy's we have \[22(x+20)=\sqrt{x^2-400}\sqrt{x^2-484}\]. We square both sides to get \[484(x+20)^2=(x^2-400)(x^2-484) \implies 484(x+20)=(x-20)(x^2-484) \implies 484x=x^3-20x^2-484x \implies x(x^2-20x-968)=0\]

We solve for $x$ via the Quadratic Formula and receive $x=10+2\sqrt{267}$, but we must divide by $2$ since we want the radius, and hence $267+5=\boxed{272}.$ ~SirAppel

See Also

2013 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_8&oldid=223176"