AIME 2008 II · 第 11 题

Solution

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

$$\frac{4}{3}r + 8\sqrt{r}+ \frac{16}{3} = 56 \rightarrow r = 44 - 6\sqrt{35} \rightarrow 44 + 6\cdot 35 = \boxed{254}$$