AMC12 2018 A

AMC12 2018 A · Q20

AMC12 2018 A · Q20. It mainly tests Triangles (properties), Circle theorems.

Triangle ABC is an isosceles right triangle with AB = AC = 3. Let M be the midpoint of hypotenuse BC. Points I and E lie on sides AC and AB, respectively, so that AI > AE and AIME is a cyclic quadrilateral. Given that triangle EMI has area 2, the length CI can be written as $a-\frac{\sqrt{b}}{c}$, where a, b, and c are positive integers and b is square-free. What is the value of a + b + c?

三角形 ABC 是等腰直角三角形，$AB = AC = 3$。$M$ 为斜边 BC 中点。点 I 和 E 在边 AC 和 AB 上，且 $AI > AE$，使得 AIME 是圆内接四边形。已知三角形 EMI 面积为 2，$CI$ 的长度可写为 $a-\frac{\sqrt{b}}{c}$，其中 $a, b, c$ 为正整数且 $b$ 无平方因子。求 $a + b + c$ 的值？

(A) 9 9

(B) 10 10

(D) 12 12

(E) 13 13

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): It follows from the Pythagorean Theorem that $CM = MB = \frac{3}{2}\sqrt{2}$. Because quadrilateral $AIME$ is cyclic, opposite angles are supplementary and thus $\angle IME$ is a right angle. Let $x = CI$ and $y = BE$; then $AI = 3 - x$ and $AE = 3 - y$. By the Law of Cosines in $\triangle MCI$, $$ IM^2 = x^2 + \left(\frac{3}{2}\sqrt{2}\right)^2 - 2\cdot x \cdot \frac{3}{2}\sqrt{2}\cdot \cos 45^\circ = x^2 - 3x + \frac{9}{2}. $$ Similarly, $ME^2 = y^2 - 3y + \frac{9}{2}$. By the Pythagorean Theorem in right triangles $EMI$ and $IAE$, $$ \left(x^2 - 3x + \frac{9}{2}\right) + \left(y^2 - 3y + \frac{9}{2}\right) = (3 - x)^2 + (3 - y)^2, $$ which simplifies to $x + y = 3$. Because the area of $\triangle EMI$ is $2$, it follows that $IM^2\cdot ME^2 = 16$. Therefore $$ \left(x^2 - 3x + \frac{9}{2}\right)\left((3 - x)^2 - 3(3 - x) + \frac{9}{2}\right)=16, $$ which simplifies to $$ \left(x^2 - 3x + \frac{9}{2}\right)^2 = 16. $$ Because $y > x$, the only real solution is $x = \frac{3-\sqrt{7}}{2}$. The requested sum is $3 + 7 + 2 = 12$.

答案（D）：由勾股定理可得 $CM = MB = \frac{3}{2}\sqrt{2}$。因为四边形 $AIME$ 为圆内接四边形，对角互补，因此 $\angle IME$ 为直角。令 $x = CI$，$y = BE$；则 $AI = 3 - x$，$AE = 3 - y$。在 $\triangle MCI$ 中由余弦定理， $$ IM^2 = x^2 + \left(\frac{3}{2}\sqrt{2}\right)^2 - 2\cdot x \cdot \frac{3}{2}\sqrt{2}\cdot \cos 45^\circ = x^2 - 3x + \frac{9}{2}. $$ 同理，$ME^2 = y^2 - 3y + \frac{9}{2}$。在直角三角形 $EMI$ 和 $IAE$ 中由勾股定理， $$ \left(x^2 - 3x + \frac{9}{2}\right) + \left(y^2 - 3y + \frac{9}{2}\right) = (3 - x)^2 + (3 - y)^2, $$ 化简得 $x + y = 3$。因为 $\triangle EMI$ 的面积为 $2$，可得 $IM^2\cdot ME^2 = 16$。因此 $$ \left(x^2 - 3x + \frac{9}{2}\right)\left((3 - x)^2 - 3(3 - x) + \frac{9}{2}\right)=16, $$ 化简为 $$ \left(x^2 - 3x + \frac{9}{2}\right)^2 = 16. $$ 因为 $y > x$，唯一的实数解为 $x = \frac{3-\sqrt{7}}{2}$。所求的和为 $3 + 7 + 2 = 12$。

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