AIME 2001 II · 第 7 题

Solution

Solution 1 (analytic)

Let $P = (0,0)$ be at the origin. Using the formula $A = rs$ on $\triangle PQR$, where $r_{1}$ is the inradius (similarly define $r_2, r_3$ to be the radii of $C_2, C_3$), $s = \frac{PQ + QR + RP}{2} = 180$ is the semiperimeter, and $A = \frac 12 bh = 5400$ is the area, we find $r_{1} = \frac As = 30$. Or, the inradius could be directly by using the formula $\frac{a+b-c}{2}$, where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. (This formula should be used only for right triangles.) Thus $ST, UV$ lie respectively on the lines $y = 60, x = 60$, and so $RS = 60, UQ = 30$.

Note that $\triangle PQR \sim \triangle STR \sim \triangle UQV$. Since the ratio of corresponding lengths of similar figures are the same, we have

$$\frac{r_{1}}{PR} = \frac{r_{2}}{RS} \Longrightarrow r_{2} = 15\ \text{and} \ \frac{r_{1}}{PQ} = \frac{r_{3}}{UQ} \Longrightarrow r_{3} = 10.$$ Let the centers of $\odot C_2, C_3$ be $O_2 = (0 + r_{2}, 60 + r_{2}) = (15, 75), O_3 = (60 + r_{3}, 0 + r_{3}) = (70,10)$, respectively; then by the distance formula we have $O_2O_3 = \sqrt{55^2 + 65^2} = \sqrt{10 \cdot 725}$. Therefore, the answer is $n = \boxed{725}$.

Solution 2 (synthetic)

We compute $r_1 = 30, r_2 = 15, r_3 = 10$ as above. Let $A_1, A_2, A_3$ respectively the points of tangency of $C_1, C_2, C_3$ with $QR$.

By the Two Tangent Theorem, we find that $A_{1}Q = 60$, $A_{1}R = 90$. Using the similar triangles, $RA_{2} = 45$, $QA_{3} = 20$, so $A_{2}A_{3} = QR - RA_2 - QA_3 = 85$. Thus $(O_{2}O_{3})^{2} = (15 - 10)^{2} + (85)^{2} = 7250\implies n=\boxed{725}$.

Solution 3

The radius of an incircle is $r=A_t/\text{semiperimeter}$. The area of the triangle is equal to $\frac{90\times120}{2} = 5400$ and the semiperimeter is equal to $\frac{90+120+150}{2} = 180$. The radius, therefore, is equal to $\frac{5400}{180} = 30$. Thus using similar triangles the dimensions of the triangle circumscribing the circle with center $C_2$ are equal to $120-2(30) = 60$, $\frac{1}{2}(90) = 45$, and $\frac{1}{2}\times150 = 75$. The radius of the circle inscribed in this triangle with dimensions $45\times60\times75$ is found using the formula mentioned at the very beginning. The radius of the incircle is equal to $15$.

Defining $P$ as $(0,0)$, $C_2$ is equal to $(60+15,15)$ or $(75,15)$. Also using similar triangles, the dimensions of the triangle circumscribing the circle with center $C_3$ are equal to $90-2(30)$, $\frac{1}{3}\times120$, $\frac{1}{3}\times150$ or $30,40,50$. The radius of $C_3$ by using the formula mentioned at the beginning is $10$. Using $P$ as $(0,0)$, $C_3$ is equal to $(10, 60+10)$ or $(10,70)$. Using the distance formula, the distance between $C_2$ and $C_3$: $\sqrt{(75-10)^2 +(15-70)^2}$ this equals $\sqrt{7250}$ or $\sqrt{725\times10}$, thus $n$ is $\boxed{725}$.

Note

The problem can be reduced to a $3-4-5$ triangle for the initial calculations, where $C_2$ is calculated to be ($\frac{5}{2}, \frac{1}{2})$, and $C_3$ is calculated to be ($\frac{1}{3}, \frac{7}{3})$. After we find the incenters the points can be scaled up by a factor of $30$ for the final distance calculation.

Solution 4 (easy but hard to see)

We can calculate the inradius of each triangle as with the previous solutions. Now, notice that the pentagon $PSTVU$ is a square with its corner cut off. We literally complete the square and mark the last corner as point X. Now, construct right triangle with $C_3C_2$ as its hypotenuse. The right angle will be at point $Y$. We will now find the length of each leg and use the Pythagorean Theorem to compute the desired length. We see that the length of $C_3Y$ is $50 + 15 = 65$, as seen by the inradius of $C_2$ and $10$ less than the square's side length. $C_2Y$ is $45 + 10 = 55$, which is $15$ less than the square plus the inradius of $C_3$. Our final answer is $\sqrt{65^2 + 55^2} = \sqrt{7250} = \sqrt{(10)(725)} \rightarrow \boxed{725}$.