Difference between revisions of "2015 AIME II Problems/Problem 15"
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==Solution== | ==Solution== | ||
+ | <asy> | ||
+ | unitsize(35); | ||
+ | draw(Circle((-1,0),1)); | ||
+ | draw(Circle((4,0),4)); | ||
+ | pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; | ||
+ | A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); | ||
+ | label("$A$",A,NE);label("$O_1$",O_1,NE);label("$O_2$",O_2,NE);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,NE);label("$N$",N,W);label("$K$",(-24/15,0.2));label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4)); | ||
+ | draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); | ||
+ | dot(O_1);dot(O_2); | ||
+ | draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); | ||
+ | </asy> | ||
+ | |||
+ | Call <math>O_1</math> and <math>O_2</math> the centers of circles <math>\mathcal{P}</math> and <math>\mathcal{Q}</math>, respectively, and call <math>K</math> and <math>L</math> the feet of the altitudes from <math>B</math> to <math>O_1N</math> and <math>C</math> to <math>O_2N</math>, respectively. Extend <math>CB</math> and <math>O_2O_1</math> to meet at point <math>N</math>. Using the fact that <math>\triangle{O_1BN} \sim \triangle{O_2CN}</math> and setting <math>NO_1 = k</math>, we have that <math>\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}</math>. We can do some more length chasing using triangles similar to <math>OBN</math> to get that <math>AK = AL = \frac{24}{15}</math>, <math>BK = \frac{12}{15}</math>, and <math>CL = \frac{48}{15}</math>. Now, consider the circles <math>\mathcal{P}</math> and <math>\mathcal{Q}</math> on the coordinate plane, where <math>A</math> is the origin. If the line <math>\ell</math> through <math>A</math> intersects <math>\mathcal{P}</math> at <math>D</math> and <math>\mathcal{Q}</math> at <math>E</math>, then <math>4 \cdot DA = AE</math>. To verify this, notice that <math>\triangle{AO_1D} \sim \triangle{EO_2A}</math> from the fact that both triangles are isosceles with <math>\angle{O_1AD} \cong \angle{O_2AE}</math>, which are corresponding angles. Since <math>O_2A = 4\cdot O_1A</math>, we can conclude that <math>\triangle{AO_1D} \sim \triangle{EO_2A}</math>. | ||
+ | |||
+ | |||
+ | Hence, we need to find the slope <math>m</math> of line <math>\ell</math> such that the perpendicular distance <math>n</math> from <math>B</math> to <math>AD</math> is four times the perpendicular distance <math>p</math> from <math>C</math> to <math>AE</math>. This will mean that the product of the bases and heights of triangles <math>ACE</math> and <math>DBA</math> will be equal, which in turn means that their areas will be equal. Let the line <math>\ell</math> have the equation <math>y = -mx \implies m + yx = 0</math>. Then, the coordinates of <math>B</math> are <math>\left(\frac{-24}{15}, \frac{-12}{15}\right)</math>, and the coordinates of <math>C</math> are <math>\left(\frac{24}{15}, \frac{-48}{15}\right)</math>. Using the point-to-line distance formula and the fact that <math>n = 4p</math>, we have <cmath>\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}</cmath> <cmath>\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.</cmath> Since <math>m</math> takes on a positive value, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have <cmath>\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.</cmath> Thus, the equation of <math>\ell</math> is <math>y = -\frac{3}{2}x</math>. | ||
+ | |||
+ | |||
+ | The we can find the coordinates of <math>D</math> by finding the point <math>(x,y)</math> other than <math>A = (0,0)</math> where the circle <math>\mathcal{P}</math> intersects <math>\ell</math>. <math>\mathcal{P}</math> can be represented with the equation <math>(x + 1)^2 + y^2 = 1</math>, and substituting <math>y = -\frac{3}{2}x</math> into this equation yields <cmath>x^2 + 2x + 1 + \frac{9}{4}x^2 = 1 \implies \frac{13}{4}x^2 + 2x = 0 \implies x = 0, -\frac{8}{13}.</cmath> Discarding <math>x = 0</math>, the <math>y</math>-coordinate of <math>D</math> is <math>-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}</math>. The distance from <math>D</math> to <math>A</math> is then <math>\sqrt{\left(\frac{-8}{13}\right)^2 + \left(\frac{12}{13}\right)^2} = \frac{4}{\sqrt{13}}.</math> The perpendicular distance from <math>B</math> to <math>AD</math> or the height of <math>\triangle{DBA}</math> is <math>\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.</math> Finally, the common area is <math>\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}</math>, and <math>m + n = 64 + 65 = \boxed{129}</math>. | ||
==See also== | ==See also== | ||
{{AIME box|year=2015|n=II|num-b=14|after=Last Problem}} | {{AIME box|year=2015|n=II|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 05:44, 28 March 2015
Problem
Circles and
have radii
and
, respectively, and are externally tangent at point
. Point
is on
and point
is on
so that line
is a common external tangent of the two circles. A line
through
intersects
again at
and intersects
again at
. Points
and
lie on the same side of
, and the areas of
and
are equal. This common area is
, where
and
are relatively prime positive integers. Find
.
Solution
Call and
the centers of circles
and
, respectively, and call
and
the feet of the altitudes from
to
and
to
, respectively. Extend
and
to meet at point
. Using the fact that
and setting
, we have that
. We can do some more length chasing using triangles similar to
to get that
,
, and
. Now, consider the circles
and
on the coordinate plane, where
is the origin. If the line
through
intersects
at
and
at
, then
. To verify this, notice that
from the fact that both triangles are isosceles with
, which are corresponding angles. Since
, we can conclude that
.
Hence, we need to find the slope of line
such that the perpendicular distance
from
to
is four times the perpendicular distance
from
to
. This will mean that the product of the bases and heights of triangles
and
will be equal, which in turn means that their areas will be equal. Let the line
have the equation
. Then, the coordinates of
are
, and the coordinates of
are
. Using the point-to-line distance formula and the fact that
, we have
Since
takes on a positive value, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have
Thus, the equation of
is
.
The we can find the coordinates of by finding the point
other than
where the circle
intersects
.
can be represented with the equation
, and substituting
into this equation yields
Discarding
, the
-coordinate of
is
. The distance from
to
is then
The perpendicular distance from
to
or the height of
is
Finally, the common area is
, and
.
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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