AMC10 2019 B

AMC10 2019 B · Q15

AMC10 2019 B · Q15. It mainly tests Quadratic equations, Pythagorean theorem.

Right triangles $T_1$ and $T_2$ have areas 1 and 2, respectively. A side of $T_1$ is congruent to a side of $T_2$, and a different side of $T_1$ is congruent to a different side of $T_2$. What is the square of the product of the lengths of the other (third) sides of $T_1$ and $T_2$?

直角三角形$T_1$和$T_2$的面积分别是1和2。$T_1$的一条边与$T_2$的一条边全等，$T_1$的另一条不同的边与$T_2$的另一条不同的边全等。$T_1$和$T_2$其余（第三）边的长度乘积的平方是多少？

(A) \frac{28}{3} \frac{28}{3}

(B) 10 10

(D) \frac{32}{3} \frac{32}{3}

(E) 12 12

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Let $a$ and $b$, with $a<b$, be the shared side lengths. Then $T_1$ has hypotenuse $b$ and legs $a$ and $\sqrt{b^2-a^2}$, and $T_2$ has hypotenuse $\sqrt{a^2+b^2}$ and legs $a$ and $b$. Thus $\frac{1}{2}a\sqrt{b^2-a^2}=1$ and $\frac{1}{2}ab=2$. Multiplying the first equation by 2 and then squaring gives $a^2b^2-a^4=4$. From the second equation, $a^2b^2=16$, so $16-a^4=4$, which means $a^4=12$. Then $$ b^4=\left(\frac{4}{a}\right)^4=\frac{4^4}{a^4}=\frac{256}{12}=\frac{64}{3}. $$ Therefore the square of the product of the other sides is $$ \left(\sqrt{b^2-a^2}\cdot\sqrt{a^2+b^2}\right)^2=b^4-a^4=\frac{64}{3}-12=\frac{28}{3}. $$

答案（A）：设共享边长为 $a$ 和 $b$，且 $a<b$。则 $T_1$ 的斜边为 $b$，两直角边为 $a$ 与 $\sqrt{b^2-a^2}$；$T_2$ 的斜边为 $\sqrt{a^2+b^2}$，两直角边为 $a$ 与 $b$。因此 $\frac{1}{2}a\sqrt{b^2-a^2}=1$ 且 $\frac{1}{2}ab=2$。将第一个等式乘以 2 再平方得 $a^2b^2-a^4=4$。由第二个等式得 $a^2b^2=16$，所以 $16-a^4=4$，从而 $a^4=12$。于是 $$ b^4=\left(\frac{4}{a}\right)^4=\frac{4^4}{a^4}=\frac{256}{12}=\frac{64}{3}. $$ 因此其余边的乘积的平方为 $$ \left(\sqrt{b^2-a^2}\cdot\sqrt{a^2+b^2}\right)^2=b^4-a^4=\frac{64}{3}-12=\frac{28}{3}. $$

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