Difference between revisions of "2008 AIME II Problems/Problem 11"
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By the [[Pythagorean Theorem]], simplification, and the [[quadratic formula]], we can get <math>r = 44 - 6\sqrt {35}</math>, for a final answer of <math>\fbox{254}</math>. | By the [[Pythagorean Theorem]], simplification, and the [[quadratic formula]], we can get <math>r = 44 - 6\sqrt {35}</math>, for a final answer of <math>\fbox{254}</math>. | ||
+ | |||
+ | == Solution 2(pure synthetic) == | ||
+ | Refer to the above diagram. Let the larger circle have center <math>O_1</math>, the smaller have center <math>O_2</math>, and the incenter be <math>I</math>. We can easily calculate that the area of <math>\triangle ABC = 2688</math>, and <math>s = 128</math> and <math>R = 21</math>, where <math>R</math> is the inradius. | ||
+ | |||
+ | Now, Line <math>\overline{AI}</math> is the perpendicular bisector of <math>\overline{BC}</math>, as <math>\triangle ABC</math> is isosceles. Letting the point of intersection be <math>X</math>, we get that <math>BX = 28</math> and <math>IX = 21</math>, and <math>B, O_2, I</math> are collinear as <math>O_2</math> is equidistant from <math>\overline{AB}</math> and <math>\overline{BC}</math>. By Pythagoras, <math>BI = 35</math>, and we notice that <math>\triangle BIX</math> is a 3-4-5 right triangle. | ||
+ | |||
+ | Letting <math>r</math> be the desired radius and letting <math>Y</math> be the projection of <math>O_2</math> onto <math>\overline{BC}</math>, we find that <math>BY = \frac{4r}{3}</math>. Similarly, we find that the distance between the projection from <math>O_1</math> onto <math>\overline{BC}</math>, <math>W</math>, and <math>C</math>, is <math>\frac{64}{3}</math>. From there, we let the projection of <math>O_2</math> onto <math>\overline{O_1W}</math> be <math>Z</math>, and we have <math>O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}</math>, <math>O_1Z = 16 - r</math>, and <math>O_1O_2 = 16 + r</math>. We finish with Pythagoras on <math>\triangle O_1O_2Z</math>, whence we get the desired answer of <math>\boxed{254}</math>. - Spacesam | ||
== See also == | == See also == |
Revision as of 20:32, 30 August 2019
Problem
In triangle ,
, and
. Circle
has radius
and is tangent to
and
. Circle
is externally tangent to
and is tangent to
and
. No point of circle
lies outside of
. The radius of circle
can be expressed in the form
, where
,
, and
are positive integers and
is the product of distinct primes. Find
.
Solution
![[asy] size(200); pathpen=black;pointpen=black;pen f=fontsize(9); real r=44-6*35^.5; pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P); path PC=CR(P,16),QC=CR(Q,r); D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed); D(PC); D(QC); MP("A",A,N,f);MP("B",B,f);MP("C",C,f);MP("X",X,f);MP("Y",Y,f);D(MP("P",P,NW,f));D(MP("Q",Q,NW,f)); [/asy]](http://latex.artofproblemsolving.com/3/e/d/3edb3ca51a589034396103f5202cb9018c7289d9.png)
Let and
be the feet of the perpendiculars from
and
to
, respectively. Let the radius of
be
. We know that
. From
draw segment
such that
is on
. Clearly,
and
. Also, we know
is a right triangle.
To find , consider the right triangle
. Since
is tangent to
, then
bisects
. Let
; then
. Dropping the altitude from
to
, we recognize the
right triangle, except scaled by
.
So we get that . From the half-angle identity, we find that
. Therefore,
. By similar reasoning in triangle
, we see that
.
We conclude that .
So our right triangle has sides
,
, and
.
By the Pythagorean Theorem, simplification, and the quadratic formula, we can get , for a final answer of
.
Solution 2(pure synthetic)
Refer to the above diagram. Let the larger circle have center , the smaller have center
, and the incenter be
. We can easily calculate that the area of
, and
and
, where
is the inradius.
Now, Line is the perpendicular bisector of
, as
is isosceles. Letting the point of intersection be
, we get that
and
, and
are collinear as
is equidistant from
and
. By Pythagoras,
, and we notice that
is a 3-4-5 right triangle.
Letting be the desired radius and letting
be the projection of
onto
, we find that
. Similarly, we find that the distance between the projection from
onto
,
, and
, is
. From there, we let the projection of
onto
be
, and we have
,
, and
. We finish with Pythagoras on
, whence we get the desired answer of
. - Spacesam
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.