AMC12 2014 B

AMC12 2014 B · Q24

AMC12 2014 B · Q24. It mainly tests Quadratic equations, Circle theorems.

Let $ABCDE$ be a pentagon inscribed in a circle such that $AB = CD = 3$, $BC = DE = 10$, and $AE = 14$. The sum of the lengths of all diagonals of $ABCDE$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?

设$ABCDE$是一个内接于圆的五边形，使得$AB = CD = 3$，$BC = DE = 10$，且$AE = 14$。五边形$ABCDE$所有对角线长度之和等于$\frac{m}{n}$，其中$m$和$n$互质。求$m + n$。

(A) 129 129

(B) 247 247

(D) 391 391

(E) 421 421

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $x=AC$, $y=AD$, and $z=BE$. Because the arcs $ABC$, $BCD$, and $CDE$ are congruent, it follows that $AC=BD=CE=x$. By Ptolemy’s Theorem applied to the quadrilaterals $ABCD$, $ABDE$, and $BCDE$, it follows that $$10y+9=x^2,\quad 30+14x=yz,\quad \text{and}\quad 100+3z=x^2.$$ Solving for $y$ and $z$ in the first and third equations and substituting in the second equation gives $$30+14x=\left(\frac{x^2-9}{10}\right)\left(\frac{x^2-100}{3}\right)=\frac{x^4-109x^2+900}{30},$$ which implies that $$900+420x=x^4-109x^2+900.$$ Thus $x^3-109x-420=0$. This equation factors as $(x-12)(x+5)(x+7)=0$. Because $x>0$ it follows that $x=12$, $y=\frac{1}{10}(x^2-9)=\frac{135}{10}=\frac{27}{2}$, and $z=\frac{1}{3}(x^2-100)=\frac{44}{3}$. The required sum of diagonals equals $3x+y+z=\frac{385}{6}$, so $m+n=385+6=391$.

答案（D）：设 $x=AC$，$y=AD$，$z=BE$。由于弧 $ABC$、$BCD$、$CDE$ 全等，可得 $AC=BD=CE=x$。对四边形 $ABCD$、$ABDE$、$BCDE$ 应用托勒密定理，得到 $$10y+9=x^2,\quad 30+14x=yz,\quad \text{且}\quad 100+3z=x^2.$$ 由第一、第三个方程解出 $y$ 和 $z$，代入第二个方程，得 $$30+14x=\left(\frac{x^2-9}{10}\right)\left(\frac{x^2-100}{3}\right)=\frac{x^4-109x^2+900}{30},$$ 从而 $$900+420x=x^4-109x^2+900.$$ 因此 $x^3-109x-420=0$。因式分解为 $(x-12)(x+5)(x+7)=0$。由于 $x>0$，可得 $x=12$，$y=\frac{1}{10}(x^2-9)=\frac{135}{10}=\frac{27}{2}$，以及 $z=\frac{1}{3}(x^2-100)=\frac{44}{3}$。所求对角线之和为 $3x+y+z=\frac{385}{6}$，所以 $m+n=385+6=391$。

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