AMC12 2007 B

AMC12 2007 B · Q23

AMC12 2007 B · Q23. It mainly tests Quadratic equations, Divisibility & factors.

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?

有多少个两条直角边长为正整数且互不全等的直角三角形，使得它们的面积在数值上等于其周长的 $3$ 倍？

(A) 6 6

(B) 7 7

(D) 10 10

(E) 12 12

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $a$ and $b$ be the two legs of the triangle. We have $\frac{1}{2}ab = 3(a+b+c)$. Then $ab=6 \left(a+b+\sqrt {a^2 + b^2}\right)$. We can complete the square under the root, and we get, $ab=6 \left(a+b+\sqrt {(a+b)^2 - 2ab}\right)$. Let $ab=p$ and $a+b=s$, we have $p=6 \left(s+ \sqrt {s^2 - 2p}\right)$. After rearranging, squaring both sides, and simplifying, we have $p=12s-72$. Putting back $a$ and $b$, and after factoring using Simon's Favorite Factoring Trick, we've got $(a-12)(b-12)=72$. Factoring 72, we get 6 pairs of $a$ and $b$ $(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).$ And this gives us $6$ solutions $\Rightarrow \mathrm{(A)}$. Alternatively, note that $72 = 2^3 \cdot 3^2$. Then 72 has $(3+1)(2+1) = (4)(3) = 12$ factors. However, half of these are repeats, so we have $\frac{12}{2} = 6$ solutions.

设两条直角边为 $a$ 和 $b$。有 $\frac{1}{2}ab = 3(a+b+c)$。于是 $ab=6 \left(a+b+\sqrt {a^2 + b^2}\right)$。将根号内配方，得 $ab=6 \left(a+b+\sqrt {(a+b)^2 - 2ab}\right)$。令 $ab=p$ 且 $a+b=s$，则 $p=6 \left(s+ \sqrt {s^2 - 2p}\right)$。移项、两边平方并化简，得到 $p=12s-72$。代回 $a$ 与 $b$，并用 Simon's Favorite Factoring Trick 分解，得到 $(a-12)(b-12)=72$。分解 72，可得 6 组 $(a,b)$： $(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).$ 因此共有 $6$ 个解 $\Rightarrow \mathrm{(A)}$。或者注意到 $72 = 2^3 \cdot 3^2$，所以 72 有 $(3+1)(2+1) = (4)(3) = 12$ 个因子。但其中一半是重复的，因此有 $\frac{12}{2} = 6$ 个解。

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