Difference between revisions of "2008 AIME II Problems/Problem 11"
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pathpen=black;pointpen=black;pen f=fontsize(9); | pathpen=black;pointpen=black;pen f=fontsize(9); | ||
real r=44-6*35^.5; | 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, | + | 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); | 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(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed); | ||
Line 25: | Line 25: | ||
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 == | ||
+ | First let <math>\theta = \angle{PCB}</math> ; now connect the points as shown in the first solution's diagram ; realise that <math>\tan\theta = r/x = 16/y = r + 16/(x+y)</math> where <math>x = BY</math> and <math>y = CX</math> (the 2 tangents) ; then we have that <math>QM = 64r = 56 - x - y \implies (x+y) = 56 - 64r</math> | ||
+ | hence <math>r/x = 16+r/(56-64r)</math> ; now drop altitude <math>AY</math> to solve for <math>\tan{2\theta}</math> ; now since we know <math>\tan{2\theta}</math> we know <math>\tan \theta = r/x</math> in terms of <math>r</math> hence solve the resulting equation in <math>r</math>. | ||
+ | |||
+ | == Solution 3 (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 | ||
+ | |||
+ | == Solution 4 == | ||
+ | Let the incenter be O and the altitude from A to <math>\overline{BC}</math> be T. Note that by AA, <math>\triangle BQY \sim \triangle OBT</math> and <math>\triangle PXC \sim \triangle OTC.</math> Note that from <math>A = rs</math>, the inradius of the big triangle is <math>21</math> Using ravi substitution(or noticing that <math>\overline{AT}</math> is an altitude), we then have that <math>TB = TC = 28.</math> From similar triangles, we can now find <math>\overline{BY}.</math> We have <cmath>\frac{\overline{BY}}{QY} = \frac{7}{{21}} \rightarrow \overline{BY} = \frac{4}{3} r</cmath> Now, note that as in solution 1, drawing the perpendicular from Q to <math>\overline{PX}</math>(call it Z) yields <math>\overline{PZ} = 16 - r, \overline{ZX} = r.</math> Then, from this, <cmath>\overline{QZ} = \overline{YX} = \sqrt{(\overline{PQ})^2 - (\overline{PZ})^2} = \sqrt{(16+r)^2-(16-r)^2} = 8\sqrt{r}</cmath> Using similar similarity as was done to find <math>\overline{BY}</math> we have <math>\frac{\overline{PX}}{\overline{XC}} = \frac{\overline{OT}}{\overline{TC}} \rightarrow \frac{16}{\overline{XC}} = \frac{21}{28} \rightarrow \overline{XC} = \frac{64}{3}</math>. | ||
+ | Now adding all these up and equating them to <math>\overline{BC}</math> yields | ||
+ | <cmath>\frac{4}{3}r + 8\sqrt{r}+ \frac{16}{3} = 56 \rightarrow r = 44 - 6\sqrt{35} \rightarrow 44 + 6\cdot 35 = \boxed{254}</cmath> | ||
== See also == | == See also == |
Latest revision as of 22:22, 18 August 2024
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
First let ; now connect the points as shown in the first solution's diagram ; realise that
where
and
(the 2 tangents) ; then we have that
hence
; now drop altitude
to solve for
; now since we know
we know
in terms of
hence solve the resulting equation in
.
Solution 3 (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
Solution 4
Let the incenter be O and the altitude from A to be T. Note that by AA,
and
Note that from
, the inradius of the big triangle is
Using ravi substitution(or noticing that
is an altitude), we then have that
From similar triangles, we can now find
We have
Now, note that as in solution 1, drawing the perpendicular from Q to
(call it Z) yields
Then, from this,
Using similar similarity as was done to find
we have
.
Now adding all these up and equating them to
yields
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.