Difference between revisions of "2012 AIME I Problems/Problem 13"
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− | ==Problem | + | ==Problem== |
Three concentric circles have radii <math>3,</math> <math>4,</math> and <math>5.</math> An equilateral triangle with one vertex on each circle has side length <math>s.</math> The largest possible area of the triangle can be written as <math>a + \tfrac{b}{c} \sqrt{d},</math> where <math>a,</math> <math>b,</math> <math>c,</math> and <math>d</math> are positive integers, <math>b</math> and <math>c</math> are relatively prime, and <math>d</math> is not divisible by the square of any prime. Find <math>a+b+c+d.</math> | Three concentric circles have radii <math>3,</math> <math>4,</math> and <math>5.</math> An equilateral triangle with one vertex on each circle has side length <math>s.</math> The largest possible area of the triangle can be written as <math>a + \tfrac{b}{c} \sqrt{d},</math> where <math>a,</math> <math>b,</math> <math>c,</math> and <math>d</math> are positive integers, <math>b</math> and <math>c</math> are relatively prime, and <math>d</math> is not divisible by the square of any prime. Find <math>a+b+c+d.</math> | ||
− | + | ||
− | + | ==Solution 1== | |
− | Reinterpret the problem in the following manner. Equilateral triangle <math>ABC</math> has a point <math>X</math> on the interior such that <math>AX = 5,</math> <math>BX = 4,</math> and <math>CX = 3.</math> A <math>60^\circ</math> counter-clockwise rotation about vertex <math>A</math> maps <math>X</math> to <math>X'</math> and <math> | + | Reinterpret the problem in the following manner. Equilateral triangle <math>ABC</math> has a point <math>X</math> on the interior such that <math>AX = 5,</math> <math>BX = 4,</math> and <math>CX = 3.</math> A <math>60^\circ</math> counter-clockwise rotation about vertex <math>A</math> maps <math>X</math> to <math>X'</math> and <math>B</math> to <math>C.</math> |
+ | <center><asy>import cse5; | ||
+ | size(200); | ||
+ | defaultpen(linewidth(0.4)+fontsize(8)); | ||
+ | |||
+ | pair O = origin; | ||
+ | pair A,B,C,Op,Bp,Cp; | ||
+ | path c3,c4,c5; | ||
+ | c3 = CR(O,3); | ||
+ | c4 = CR(O,4); | ||
+ | c5 = CR(O,5); | ||
+ | draw(c3^^c4^^c5, gray+0.25); | ||
+ | A = 5*dir(96.25); | ||
+ | Op = rotate(60,A)*O; | ||
+ | B = OP(CR(Op,4),c3); | ||
+ | Bp = IP(CR(Op,4),c3); | ||
+ | C = rotate(-60,A)*B; | ||
+ | Cp = rotate(-60,A)*Bp; | ||
+ | draw(A--B--C--A, black+0.8); | ||
+ | draw(A--Bp--Cp--A, royalblue+0.8); | ||
+ | draw(CR(Op,4), heavygreen+0.25); | ||
+ | dot("$A$",A,N); | ||
+ | dot("$X$",O,E); | ||
+ | dot("$X'$",Op,E); | ||
+ | dot("$C$",B,SE); | ||
+ | dot("$C'$",Bp,NE); | ||
+ | dot("$B$",C,2*SW); | ||
+ | dot("$B'$",Cp,2*S); | ||
+ | </asy></center> | ||
+ | Note that angle <math>XAX'</math> is <math>60</math> and <math>XA = X'A = 5</math> which tells us that triangle <math>XAX'</math> is equilateral and that <math>XX' = 5.</math> We now notice that <math>XC = 3</math> and <math>X'C = 4</math> which tells us that angle <math>XCX'</math> is <math>90</math> because there is a <math>3</math>-<math>4</math>-<math>5</math> Pythagorean triple. Now note that <math>\angle ABC + \angle ACB = 120^\circ</math> and <math>\angle XCA + \angle XBA = 90^\circ,</math> so <math>\angle XCB+\angle XBC = 30^\circ</math> and <math>\angle BXC = 150^\circ.</math> Applying the law of cosines on triangle <math>BXC</math> yields | ||
<cmath>BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos150^\circ = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}</cmath> | <cmath>BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos150^\circ = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}</cmath> | ||
− | and thus the area of <math>ABC</math> equals <cmath> | + | and thus the area of <math>ABC</math> equals <cmath>\frac{\sqrt{3}}{4}\cdot BC^2 = 25\frac{\sqrt{3}}{4}+9.</cmath> |
so our final answer is <math>3+4+25+9 = \boxed{041}.</math> | so our final answer is <math>3+4+25+9 = \boxed{041}.</math> | ||
− | + | ==Solution 2== | |
− | + | Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties. | |
− | + | We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have two cases to consider; either the center <math>O</math> of the circles lies in the interior of triangle <math>ABC</math> or it does not (and we shall show that both can happen). To see that the first case can occur proceed as follows. Using the notation from Solution 1, let <math>\angle XAC = \theta</math> so that <math>\angle BAX=60-\theta</math>. Let <math>AB=BC=AC=x</math>. The law of cosines on <math>\triangle BAX</math> and <math>\triangle CAX</math> yields <cmath>\cos(60-\theta)=\frac{x^2+9}{10x}\quad \text{ and }\quad \cos\theta=\frac{x^2+16}{10x}.</cmath> Solving this system will yield the value of <math>x</math>. Since <math>\cos\theta=\frac{x^2+16}{10x}</math> we have that <cmath>\sin\theta=\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}.</cmath> Substituting these into the equation <cmath>\frac{x^2+9}{10x}=\cos(60-\theta)=\tfrac{1}{2}(\cos\theta+ \sqrt{3}\sin\theta)</cmath> we obtain <cmath>\frac{x^2+9}{10x}=\frac{1}{2}\frac{x^2+16}{10x}+\frac{\sqrt{3}}{2}\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}.</cmath> After clearing denominators, combining like terms, isolating the square root, squaring, and expanding, we obtain <cmath>x^4-50x^2+193=0</cmath> so that by the quadratic formula <math>x^2=25\pm12\sqrt{3}</math>. Under the hypothesis that <math>O</math> lies in the interior of triangle <math>ABC</math>, <math>x^2</math> must be <math>25+12\sqrt{3}</math>. To see this, note that the other value for <math>x^2</math> is roughly <math>4.2</math> so that <math>x\approx 2.05</math>, but since <math>AX=5</math> and <math>AX\leq x</math> we have a contradiction. We then obtain the area as in Solution 1. | |
− | |||
− | We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have | ||
Now, suppose <math>O</math> does not lie in the interior of triangle <math>ABC</math>. We then obtain convex quadrilateral <math>OBAC</math> with diagonals <math>CB</math> and <math>OA</math> intersecting at <math>X</math>. Here <math>AX=AB=AC=x</math>. We may let <math>\alpha</math> denote the measure of angle <math>CAX</math> so that angle <math>XAB</math> measures <math>60-\alpha</math>. Note that the law of cosines on triangles <math>CXA</math> and <math>BXA</math> yield the same equations as in the first case with <math>\theta</math> replaced with <math>\alpha</math>. Thus we obtain again <math>x^2=25\pm12\sqrt{3}</math>. If <math>x^2=25+12\sqrt{3}</math> then <math>x\approx 6.8</math>, but this is impossible since <math>AX\leq 5</math> but the shortest possible distance from <math>A</math> to <math>X</math> is the height of equilateral triangle <math>ABC</math> which is <math>\approx6.8\sqrt{3}\approx5.8</math>; a contradiction. Hence in this case <math>x^2=25-12\sqrt{3}</math>. But, the area of this triangle is clearly less than that in the first case, so we are done. Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence). | Now, suppose <math>O</math> does not lie in the interior of triangle <math>ABC</math>. We then obtain convex quadrilateral <math>OBAC</math> with diagonals <math>CB</math> and <math>OA</math> intersecting at <math>X</math>. Here <math>AX=AB=AC=x</math>. We may let <math>\alpha</math> denote the measure of angle <math>CAX</math> so that angle <math>XAB</math> measures <math>60-\alpha</math>. Note that the law of cosines on triangles <math>CXA</math> and <math>BXA</math> yield the same equations as in the first case with <math>\theta</math> replaced with <math>\alpha</math>. Thus we obtain again <math>x^2=25\pm12\sqrt{3}</math>. If <math>x^2=25+12\sqrt{3}</math> then <math>x\approx 6.8</math>, but this is impossible since <math>AX\leq 5</math> but the shortest possible distance from <math>A</math> to <math>X</math> is the height of equilateral triangle <math>ABC</math> which is <math>\approx6.8\sqrt{3}\approx5.8</math>; a contradiction. Hence in this case <math>x^2=25-12\sqrt{3}</math>. But, the area of this triangle is clearly less than that in the first case, so we are done. Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence). | ||
− | + | ==Solution 3== | |
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<center><asy> | <center><asy> | ||
import olympiad; | import olympiad; | ||
Line 81: | Line 85: | ||
By Mathdummy | By Mathdummy | ||
− | + | ==Solution 4== | |
− | Let <math>ABC</math> be the equilateral triangle with <math>AB=BC=CA=x.</math> Assume the coordinates of the vertices are <math>A(-\ | + | Let <math>ABC</math> be the equilateral triangle with <math>AB=BC=CA=x.</math> Assume the coordinates of the vertices are <math>A(-\tfrac{x}{2},0), B(\tfrac{x}{2},0)</math> and <math>C(0,\tfrac{\sqrt{3}}{2}x).</math> Let <math>P(a,b)</math> be such that <math>PA=3, PB=4</math> and <math>PC=5.</math> Then |
<cmath>\left (a+\dfrac{x}{2}\right )^2 +b^2=3^2,</cmath> | <cmath>\left (a+\dfrac{x}{2}\right )^2 +b^2=3^2,</cmath> | ||
<cmath>\left (a-\dfrac{x}{2}\right )^2 +b^2=4^2,</cmath> | <cmath>\left (a-\dfrac{x}{2}\right )^2 +b^2=4^2,</cmath> | ||
Line 91: | Line 95: | ||
-JZ | -JZ | ||
+ | ==Solution 5== | ||
+ | |||
+ | <asy> | ||
+ | import cse5; size(350); defaultpen(linewidth(0.6)+fontsize(12)); | ||
+ | pair O=origin; pair Op,Bp,Cp; | ||
+ | path c3 = CR(O,3), c4 = CR(O,4), c5 = CR(O,5); | ||
+ | var theta=55.75; | ||
+ | pair A = 3*dir(theta), Ap = rotate(150,O)*A, F=IP(c4,O--2*Ap), C=rotate(60,A)*F, E=rotate(60,A)*C, B=IP(c5,O--E), D=foot(C,O,E); | ||
+ | |||
+ | filldraw(A--O--B--cycle, rgb(255,255,200)); | ||
+ | draw(c3, cyan); | ||
+ | draw(c4, green); | ||
+ | draw(c5, purple); | ||
+ | draw(A--F--C--A--E--C, red+0.8); | ||
+ | draw(D--B--C--D--O--C, purple+0.6); | ||
+ | draw(A--O^^E--B, cyan+0.6); | ||
+ | draw(A--B^^O--F, heavygreen+0.6); | ||
+ | draw(A--D); | ||
+ | dot("$A$",A,N); dot("$O$",O,dir(A-B)); dot("$F$",F,dir(F-E)); dot("$C$",C,SE); dot("$B$",B,dir(-45)); dot("$D$",D,dir(B-C)); dot("$E$",E,dir(E-F)); | ||
+ | label("$s$",A--F,dir(A-E),red); | ||
+ | label("$s$",F--C,dir(C-B),red); | ||
+ | label("$s$",A--C,dir(C-E),red); | ||
+ | label("$s$",A--E,dir(E-C),red); | ||
+ | label("$s$",C--E,dir(C),red); | ||
+ | label("$x$",B--E,down,royalblue); | ||
+ | label("$x$",O--A,dir(B-A),royalblue); | ||
+ | label("$y$",A--B,dir(F-A),heavygreen); | ||
+ | label("$y$",O--F,dir(C),heavygreen); | ||
+ | label("$z$",C--B,dir(E-A),purple); | ||
+ | label("$z$",O--C,dir(F),purple); | ||
+ | label("$m$",A--D,dir(0),blue+fontsize(9)); | ||
+ | markscalefactor=0.03; | ||
+ | draw(rightanglemark(C,D,B), gray); | ||
+ | MA("\varphi",A,D,O,0.25,blue); | ||
+ | </asy> | ||
+ | We have <math>x=3</math>, <math>y=4</math>, and <math>z=5</math>. Because <math>AD=m</math> is the median of <math>\triangle AOB</math>, by Stewart's Theorem we have <cmath>m^2=\frac 12 (x^2+y^2)-\frac 14 \cdot z^2\quad \Rightarrow \quad m = \frac 52.</cmath> Because <math>CD</math> is the altitude of equilateral triangle <math>OBC</math>, we have <math>CD=\frac{\sqrt{3}}2\cdot z</math>. Then in <math>\triangle ADC</math>, we have <math>\angle ADC=\varphi+90^\circ</math>, so <math>\cos(\angle ADC)=-\sin\varphi</math>, and the Law of Cosines gives<cmath>s^2=m^2+\frac 34\cdot z^2 + mz\sqrt{3}\sin\varphi = 25 \left(1+\frac {\sqrt{3}}{2}\cdot\sin\varphi \right)</cmath>To calculate <math>\sin\phi</math> we apply the Law of Cosines to <math>\triangle ADO</math> to get<cmath>mz\cos\varphi = m^2+\frac 14\cdot z^2 - x^2 \quad \Rightarrow \quad \cos\varphi = \frac 7{25}\quad \Rightarrow \quad \sin\varphi = \frac {24}{25}.</cmath> | ||
+ | Finally, we get <math>s^2=25+12\sqrt{3}</math> and and thus the area of <math>\triangle ABC</math> equals<cmath>\frac{\sqrt{3}}{4}\cdot s^2 = 25\frac{\sqrt{3}}{4}+9.</cmath> | ||
+ | so our final answer is <math>3+4+25+9 = \boxed{041}.</math> | ||
+ | |||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
− | + | https://artofproblemsolving.com/videos/amc/2012aimei/353 | |
− | |||
− | + | ~ dolphin7 | |
− | |||
− | |||
− | |||
==See also== | ==See also== | ||
+ | [[2020 AMC 12A Problems/Problem 24|2020 AMC 12A Problems/Problem 24]] | ||
{{AIME box|year=2012|n=I|num-b=12|num-a=14}} | {{AIME box|year=2012|n=I|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:42, 25 January 2023
Contents
Problem
Three concentric circles have radii
and
An equilateral triangle with one vertex on each circle has side length
The largest possible area of the triangle can be written as
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime. Find
Solution 1
Reinterpret the problem in the following manner. Equilateral triangle has a point
on the interior such that
and
A
counter-clockwise rotation about vertex
maps
to
and
to
![[asy]import cse5; size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair O = origin; pair A,B,C,Op,Bp,Cp; path c3,c4,c5; c3 = CR(O,3); c4 = CR(O,4); c5 = CR(O,5); draw(c3^^c4^^c5, gray+0.25); A = 5*dir(96.25); Op = rotate(60,A)*O; B = OP(CR(Op,4),c3); Bp = IP(CR(Op,4),c3); C = rotate(-60,A)*B; Cp = rotate(-60,A)*Bp; draw(A--B--C--A, black+0.8); draw(A--Bp--Cp--A, royalblue+0.8); draw(CR(Op,4), heavygreen+0.25); dot("$A$",A,N); dot("$X$",O,E); dot("$X'$",Op,E); dot("$C$",B,SE); dot("$C'$",Bp,NE); dot("$B$",C,2*SW); dot("$B'$",Cp,2*S); [/asy]](http://latex.artofproblemsolving.com/e/6/4/e64c77c00b1f44f4fb06e26cf91f9f58cba4ae79.png)
Note that angle is
and
which tells us that triangle
is equilateral and that
We now notice that
and
which tells us that angle
is
because there is a
-
-
Pythagorean triple. Now note that
and
so
and
Applying the law of cosines on triangle
yields
and thus the area of equals
so our final answer is
Solution 2
Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties.
We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have two cases to consider; either the center of the circles lies in the interior of triangle
or it does not (and we shall show that both can happen). To see that the first case can occur proceed as follows. Using the notation from Solution 1, let
so that
. Let
. The law of cosines on
and
yields
Solving this system will yield the value of
. Since
we have that
Substituting these into the equation
we obtain
After clearing denominators, combining like terms, isolating the square root, squaring, and expanding, we obtain
so that by the quadratic formula
. Under the hypothesis that
lies in the interior of triangle
,
must be
. To see this, note that the other value for
is roughly
so that
, but since
and
we have a contradiction. We then obtain the area as in Solution 1.
Now, suppose does not lie in the interior of triangle
. We then obtain convex quadrilateral
with diagonals
and
intersecting at
. Here
. We may let
denote the measure of angle
so that angle
measures
. Note that the law of cosines on triangles
and
yield the same equations as in the first case with
replaced with
. Thus we obtain again
. If
then
, but this is impossible since
but the shortest possible distance from
to
is the height of equilateral triangle
which is
; a contradiction. Hence in this case
. But, the area of this triangle is clearly less than that in the first case, so we are done. Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence).
Solution 3
![[asy] import olympiad; import cse5; import graph; dotfactor = 2; unitsize(0.3inch); pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4); pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B; pair X = extension(A,F,D,C); pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M; dot("$C$", C, dir(0)); dot("$A$", A, dir(90));dot("$B$", B, dir(180)); dot("$D$", D, NE); dot("$E$", E, dir(90));dot("$F$", F, dir(270)); dot("$M$", M, NE); dot("$N$", N, dir(270));dot("$L$", L, NW); dot("$X$", X, dir(250)); draw(L--X); draw(M--X); draw(N--X); draw(A--B--C--cycle); draw(A--D--B); draw(B--F--C); draw(A--E--C); draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed); draw(L--M--N--cycle); [/asy]](http://latex.artofproblemsolving.com/0/d/a/0da7943d44acdd6a610940b08e776b5b24eb4e48.png)
Let's call the circle center . It has a distance of 3, 4, 5 to an equilateral triangle
. Consider
’s pedal triangle
. Since
’s antipedal triangle is equilateral,
must be the one of the isogonic centers of
. We’ll take the one inside
, i.e., the Fermat point, because it leads to larger
. Now we construct the three equilateral triangles
,
, and
, the same way the Fermat point is constructed. Then we have
. Since
is concyclic with
=4 as diameter, we have
. Similarly,
, and
. So
is a 3-4-5 right triangle with
. With some more angle chasing we get
By Law of Cosines, we have
And the area follows.
By Mathdummy
Solution 4
Let be the equilateral triangle with
Assume the coordinates of the vertices are
and
Let
be such that
and
Then
Subtraction and addition of the first two equations yield
The third equation gives
Then
We can then solve for
in terms of
and have a substitution. We have
Simplify it we have a quadratic equation for
So
The larger one leads to the solution. The smaller one relates to another equilateral triangle, as indicated in Solution 2.
-JZ
Solution 5
We have
,
, and
. Because
is the median of
, by Stewart's Theorem we have
Because
is the altitude of equilateral triangle
, we have
. Then in
, we have
, so
, and the Law of Cosines gives
To calculate
we apply the Law of Cosines to
to get
Finally, we get
and and thus the area of
equals
so our final answer is
vladimir.shelomovskii@gmail.com, vvsss
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/353
~ dolphin7
See also
2020 AMC 12A Problems/Problem 24
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.