AMC12 2018 B

AMC12 2018 B · Q21

AMC12 2018 B · Q21. It mainly tests Circle theorems, Coordinate geometry.

In $\triangle ABC$ with side lengths $AB = 13$, $AC = 12$, and $BC = 5$, let $O$ and $I$ denote the circumcenter and incenter, respectively. A circle with center $M$ is tangent to the sides $AC$ and $BC$ and to the circumcircle of $\triangle ABC$. What is the area of $\triangle MOI$?

在 $\triangle ABC$ 中，边长 $AB = 13$，$AC = 12$，$BC = 5$，令 $O$ 和 $I$ 分别表示外心和内心的位置。有一个以 $M$ 为圆心、与边 $AC$ 和 $BC$ 相切且与 $\triangle ABC$ 的外接圆相切的圆。求 $\triangle MOI$ 的面积。

(A) $\frac{5}{2}$ $\frac{5}{2}$

(B) $\frac{11}{4}$ $\frac{11}{4}$

(D) $\frac{13}{4}$ $\frac{13}{4}$

(E) $\frac{7}{2}$ $\frac{7}{2}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Place the figure on coordinate axes with coordinates $A(12,0)$, $B(0,5)$, and $C(0,0)$. The center of the circumscribed circle is the midpoint of the hypotenuse of right triangle $ABC$, so the coordinates of $O$ are $(6,\frac{5}{2})$. The radius $r$ of the inscribed circle equals the area of the triangle divided by its semiperimeter, which here is $30\div 15=2$, so the center of the inscribed circle is $I(2,2)$. Because the circle with center $M$ is tangent to both coordinate axes, its center has coordinates $(\rho,\rho)$, where $\rho$ is its radius. Let $P$ be the point of tangency of this circle and the circumscribed circle. Then $M$, $P$, and $O$ are collinear because $MP$ and $OP$ are perpendicular to the common tangent line at $P$. Thus $MO=OP-MP=\frac{13}{2}-\rho$. By the distance formula, $$ MO=\sqrt{(\rho-6)^2+\left(\rho-\frac{5}{2}\right)^2}. $$ Equating these expressions and solving for $\rho$ shows that $\rho=4$. The area of $\triangle MOI$ can now be computed using the shoelace formula: $$ \left|\frac{4\cdot\frac{5}{2}+6\cdot 2+2\cdot 4-(4\cdot 6+\frac{5}{2}\cdot 2+2\cdot 4)}{2}\right|=\frac{7}{2}. $$ Alternatively, the area can be computed as $\frac{1}{2}$ times $MI$, which by the distance formula is $$ \sqrt{(4-2)^2+(4-2)^2}=2\sqrt{2}, $$ times the distance from point $O$ to the line $MI$, whose equation is $x-y+0=0$. This last value is $$ \frac{\left|1\cdot 6+(-1)\cdot\frac{5}{2}+0\right|}{\sqrt{1^2+(-1)^2}}=\frac{7}{4}\sqrt{2}, $$ so the area is $$ \frac{1}{2}\cdot(2\sqrt{2})\cdot\frac{7}{4}\sqrt{2}=\frac{7}{2}. $$

答案（E）：将图形放在坐标轴上，取坐标 $A(12,0)$、$B(0,5)$、$C(0,0)$。外接圆的圆心是直角三角形 $ABC$ 的斜边中点，因此 $O$ 的坐标为 $(6,\frac{5}{2})$。内切圆半径 $r$ 等于三角形面积除以半周长，这里为 $30\div 15=2$，所以内切圆圆心为 $I(2,2)$。由于以 $M$ 为圆心的圆与两条坐标轴都相切，其圆心坐标为 $(\rho,\rho)$，其中 $\rho$ 为半径。设 $P$ 为该圆与外接圆的切点。则 $M$、$P$、$O$ 三点共线，因为 $MP$ 和 $OP$ 都垂直于点 $P$ 处的公切线。因此 $$ MO=OP-MP=\frac{13}{2}-\rho. $$ 由距离公式， $$ MO=\sqrt{(\rho-6)^2+\left(\rho-\frac{5}{2}\right)^2}. $$ 令两式相等并解 $\rho$，得 $\rho=4$。于是可用鞋带公式计算 $\triangle MOI$ 的面积： $$ \left|\frac{4\cdot\frac{5}{2}+6\cdot 2+2\cdot 4-(4\cdot 6+\frac{5}{2}\cdot 2+2\cdot 4)}{2}\right|=\frac{7}{2}. $$ 或者，面积也可表示为 $\frac{1}{2}$ 乘以 $MI$（由距离公式） $$ \sqrt{(4-2)^2+(4-2)^2}=2\sqrt{2}, $$ 再乘以点 $O$ 到直线 $MI$ 的距离。直线 $MI$ 的方程为 $x-y+0=0$，该距离为 $$ \frac{\left|1\cdot 6+(-1)\cdot\frac{5}{2}+0\right|}{\sqrt{1^2+(-1)^2}}=\frac{7}{4}\sqrt{2}, $$ 所以面积 $$ \frac{1}{2}\cdot(2\sqrt{2})\cdot\frac{7}{4}\sqrt{2}=\frac{7}{2}. $$

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