Difference between revisions of "2013 AIME II Problems/Problem 8"
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dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); | dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); | ||
draw(unitcircle); | draw(unitcircle); | ||
− | label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D, | + | 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> | </asy> | ||
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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> | + | 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 | + | 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>. | ||
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===Solution 4=== | ===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 | 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== | ==See Also== | ||
{{AIME box|year=2013|n=II|num-b=7|num-a=9}} | {{AIME box|year=2013|n=II|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:24, 19 July 2024
Contents
Problem 8
A hexagon that is inscribed in a circle has side lengths ,
,
,
,
, and
in that order. The radius of the circle can be written as
, where
and
are positive integers. Find
.
Solution
Solution 1
Let us call the hexagon , where
, and
.
We can just consider one half of the hexagon,
, to make matters simpler.
Draw a line from the center of the circle,
, to the midpoint of
,
. Now, draw a line from
to the midpoint of
,
. Clearly,
, because
, and
, for similar reasons. Also notice that
.
Let us call
. Therefore,
, and so
. Let us label the radius of the circle
. This means
Now we can use simple trigonometry to solve for
.
Recall that
: That means
.
Recall that
: That means
.
Let
.
Substitute to get
and
Now substitute the first equation into the second equation:
Multiplying both sides by
and reordering gives us the quadratic
Using the quadratic equation to solve, we get that
(because
gives a negative value), so the answer is
.
Solution 2
Using the trapezoid mentioned above, draw an altitude of the trapezoid passing through point
onto
at point
. Now, we can use the pythagorean theorem:
. Expanding and combining like terms gives us the quadratic
and solving for
gives
. So the solution is
.
Solution 3
Join the diameter of the circle and let the length be
. By Ptolemy's Theorem on trapezoid
,
. Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to
each. Then
Since is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right
:
From the above equations, we have:
Since the radius is half the diameter, it is , so the answer is
.
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 , after simple factorization, we can get
, it is easy to see that
are two solutions for the equation, so we can factorize that into
so we only need to find the solution for
and we can get
is the desired answer for the problem, and our answer is
.~bluesoul
Solution 6 (Trig Bash)
Let . So, we have
and
. So,
. Let
be the foot of the perpendicular from
to
. We have
. Using Pythagorean theorem on
, to get
, or
. Multiplying by
, we get
. Rearranging and simplifying, we get a quadratic in
:
which gives us
. Because
is in the form
, we know to choose the larger option, meaning
, so
and
. By inspection, we get
, so our answer is
.
~Puck_0
Solution 7
We know that is a diameter, hence
and
are right triangles. Let
, and
Hence,
is a right triangle with legs
and hypotenuse,
and
is a right triangle with legs
with hypotenuse
. By Ptolemy's we have
.
We square both sides to get
We solve for via the Quadratic Formula and receive
, but we must divide by
since we want the radius, and hence
~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.