AMC12 2013 A

AMC12 2013 A · Q12

AMC12 2013 A · Q12. It mainly tests Quadratic equations, Triangles (properties).

The angles in a particular triangle are in arithmetic progression, and the side lengths are 4, 5, and $x$. The sum of the possible values of $x$ equals $a+\sqrt{b}+\sqrt{c}$, where $a$, $b$, and $c$ are positive integers. What is $a+b+c$?

某个三角形的三个角成等差数列，三边长分别为 4、5 和 $x$。所有可能的 $x$ 的取值之和等于 $a+\sqrt{b}+\sqrt{c}$，其中 $a$、$b$、$c$ 为正整数。求 $a+b+c$。

(A) 36 36

(B) 38 38

(D) 42 42

(E) 44 44

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Let the angles of the triangle be $\alpha-\delta$, $\alpha$, and $\alpha+\delta$. Then $3\alpha=\alpha-\delta+\alpha+\alpha+\delta=180^\circ$, so $\alpha=60^\circ$. There are three cases depending on which side is opposite to the $60^\circ$ angle. Suppose that the triangle is $ABC$ with $\angle BAC=60^\circ$. Let $D$ be the foot of the altitude from $C$. The triangle $CAD$ is a $30$-$60$-$90^\circ$ triangle, so $AD=\frac12 AC$ and $CD=\frac{\sqrt3}{2}AC$. There are three cases to consider. In each case the Pythagorean Theorem can be used to solve for the unknown side. If $AB=5$, $AC=4$, and $BC=x$, then $AD=2$, $CD=2\sqrt3$, and $BD=|AB-AD|=3$. It follows that $x^2=BC^2=CD^2+BD^2=21$, so $x=\sqrt{21}$. If $AB=x$, $AC=4$, and $BC=5$, then $AD=2$, $CD=2\sqrt3$, and $BD=|AB-AD|=|x-2|$. It follows that $25=BC^2=CD^2+BD^2=12+(x-2)^2$, and the positive solution is $x=2+\sqrt{13}$. If $AB=x$, $AC=5$, and $BC=4$, then $AD=\frac52$, $CD=\frac{5\sqrt3}{2}$, and $BD=|AB-AD|=|x-\frac52|$. It follows that $16=BC^2=CD^2+BD^2=\frac{75}{4}+(x-\frac52)^2$, which has no solution because $\frac{75}{4}>16$. The sum of all possible side lengths is $2+\sqrt{13}+\sqrt{21}$. The requested sum is $2+13+21=36$.

答案（A）：设三角形的三个角分别为 $\alpha-\delta$、$\alpha$、$\alpha+\delta$。则 $3\alpha=\alpha-\delta+\alpha+\alpha+\delta=180^\circ$，所以 $\alpha=60^\circ$。根据哪一边对着 $60^\circ$ 角共有三种情况。设三角形为 $ABC$ 且 $\angle BAC=60^\circ$。令 $D$ 为从 $C$ 作高的垂足。三角形 $CAD$ 是一个 $30$-$60$-$90^\circ$ 三角形，因此 $AD=\frac12 AC$，$CD=\frac{\sqrt3}{2}AC$。需要考虑三种情况。每种情况下都可以用勾股定理求未知边。若 $AB=5$，$AC=4$，$BC=x$，则 $AD=2$，$CD=2\sqrt3$，且 $BD=|AB-AD|=3$。于是 $x^2=BC^2=CD^2+BD^2=21$，所以 $x=\sqrt{21}$。若 $AB=x$，$AC=4$，$BC=5$，则 $AD=2$，$CD=2\sqrt3$，且 $BD=|AB-AD|=|x-2|$。于是 $25=BC^2=CD^2+BD^2=12+(x-2)^2$，其正解为 $x=2+\sqrt{13}$。若 $AB=x$，$AC=5$，$BC=4$，则 $AD=\frac52$，$CD=\frac{5\sqrt3}{2}$，且 $BD=|AB-AD|=|x-\frac52|$。于是 $16=BC^2=CD^2+BD^2=\frac{75}{4}+(x-\frac52)^2$，由于 $\frac{75}{4}>16$，故无解。所有可能的边长之和为 $2+\sqrt{13}+\sqrt{21}$。题目所求的和为 $2+13+21=36$。

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