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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>.
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== Solution 2 ==
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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>
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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>.
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== Solution 3 (pure synthetic) ==
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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.
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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.
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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
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== Solution 4 ==
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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>.
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Now adding all these up and equating them to <math>\overline{BC}</math> yields
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<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 21:22, 18 August 2024

Problem

In triangle $ABC$, $AB = AC = 100$, and $BC = 56$. Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$. Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$. No point of circle $Q$ lies outside of $\triangle ABC$. The radius of circle $Q$ can be expressed in the form $m - n\sqrt {k}$, where $m$, $n$, and $k$ are positive integers and $k$ is the product of distinct primes. Find $m + nk$.

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]

Let $X$ and $Y$ be the feet of the perpendiculars from $P$ and $Q$ to $BC$, respectively. Let the radius of $\odot Q$ be $r$. We know that $PQ = r + 16$. From $Q$ draw segment $\overline{QM} \parallel \overline{BC}$ such that $M$ is on $PX$. Clearly, $QM = XY$ and $PM = 16-r$. Also, we know $QPM$ is a right triangle.

To find $XC$, consider the right triangle $PCX$. Since $\odot P$ is tangent to $\overline{AC},\overline{BC}$, then $PC$ bisects $\angle ACB$. Let $\angle ACB = 2\theta$; then $\angle PCX = \angle QBX = \theta$. Dropping the altitude from $A$ to $BC$, we recognize the $7 - 24 - 25$ right triangle, except scaled by $4$.

So we get that $\tan(2\theta) = 24/7$. From the half-angle identity, we find that $\tan(\theta) = \frac {3}{4}$. Therefore, $XC = \frac {64}{3}$. By similar reasoning in triangle $QBY$, we see that $BY = \frac {4r}{3}$.

We conclude that $XY = 56 - \frac {4r + 64}{3} = \frac {104 - 4r}{3}$.

So our right triangle $QPM$ has sides $r + 16$, $r - 16$, and $\frac {104 - 4r}{3}$.

By the Pythagorean Theorem, simplification, and the quadratic formula, we can get $r = 44 - 6\sqrt {35}$, for a final answer of $\fbox{254}$.


Solution 2

First let $\theta = \angle{PCB}$ ; now connect the points as shown in the first solution's diagram ; realise that $\tan\theta = r/x = 16/y = r + 16/(x+y)$ where $x = BY$ and $y = CX$ (the 2 tangents) ; then we have that $QM = 64r = 56 - x - y \implies (x+y) = 56 - 64r$ hence $r/x = 16+r/(56-64r)$ ; now drop altitude $AY$ to solve for $\tan{2\theta}$ ; now since we know $\tan{2\theta}$ we know $\tan \theta = r/x$ in terms of $r$ hence solve the resulting equation in $r$.

Solution 3 (pure synthetic)

Refer to the above diagram. Let the larger circle have center $O_1$, the smaller have center $O_2$, and the incenter be $I$. We can easily calculate that the area of $\triangle ABC = 2688$, and $s = 128$ and $R = 21$, where $R$ is the inradius.

Now, Line $\overline{AI}$ is the perpendicular bisector of $\overline{BC}$, as $\triangle ABC$ is isosceles. Letting the point of intersection be $X$, we get that $BX = 28$ and $IX = 21$, and $B, O_2, I$ are collinear as $O_2$ is equidistant from $\overline{AB}$ and $\overline{BC}$. By Pythagoras, $BI = 35$, and we notice that $\triangle BIX$ is a 3-4-5 right triangle.

Letting $r$ be the desired radius and letting $Y$ be the projection of $O_2$ onto $\overline{BC}$, we find that $BY = \frac{4r}{3}$. Similarly, we find that the distance between the projection from $O_1$ onto $\overline{BC}$, $W$, and $C$, is $\frac{64}{3}$. From there, we let the projection of $O_2$ onto $\overline{O_1W}$ be $Z$, and we have $O_2Z = 28 - \frac{4r}{3} + \frac{20}{3}$, $O_1Z = 16 - r$, and $O_1O_2 = 16 + r$. We finish with Pythagoras on $\triangle O_1O_2Z$, whence we get the desired answer of $\boxed{254}$. - Spacesam

Solution 4

Let the incenter be O and the altitude from A to $\overline{BC}$ be T. Note that by AA, $\triangle BQY \sim \triangle OBT$ and $\triangle PXC \sim \triangle OTC.$ Note that from $A = rs$, the inradius of the big triangle is $21$ Using ravi substitution(or noticing that $\overline{AT}$ is an altitude), we then have that $TB = TC = 28.$ From similar triangles, we can now find $\overline{BY}.$ We have \[\frac{\overline{BY}}{QY} = \frac{7}{{21}} \rightarrow \overline{BY} = \frac{4}{3} r\] Now, note that as in solution 1, drawing the perpendicular from Q to $\overline{PX}$(call it Z) yields $\overline{PZ} = 16 - r, \overline{ZX} = r.$ Then, from this, \[\overline{QZ} = \overline{YX} = \sqrt{(\overline{PQ})^2 - (\overline{PZ})^2} = \sqrt{(16+r)^2-(16-r)^2} = 8\sqrt{r}\] Using similar similarity as was done to find $\overline{BY}$ we have $\frac{\overline{PX}}{\overline{XC}} = \frac{\overline{OT}}{\overline{TC}} \rightarrow \frac{16}{\overline{XC}} = \frac{21}{28} \rightarrow \overline{XC} = \frac{64}{3}$. Now adding all these up and equating them to $\overline{BC}$ yields \[\frac{4}{3}r + 8\sqrt{r}+ \frac{16}{3} = 56 \rightarrow r = 44 - 6\sqrt{35} \rightarrow 44 + 6\cdot 35 = \boxed{254}\]

See also

2008 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AIME Problems and Solutions

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