AMC10 2021 A

AMC10 2021 A · Q17

AMC10 2021 A · Q17. It mainly tests Angle chasing, Triangles (properties).

Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD},BC=CD=43$, and $\overline{AD}\perp\overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. Given that $OP=11$, the length of $AD$ can be written in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$?

梯形 $ABCD$ 有 $\overline{AB}\parallel\overline{CD}$，$BC=CD=43$，且 $\overline{AD}\perp\overline{BD}$。设 $O$ 为对角线 $\overline{AC}$ 和 $\overline{BD}$ 的交点，$P$ 为 $\overline{BD}$ 的中点。已知 $OP=11$，$AD$ 的长度可以写成 $m\sqrt{n}$ 的形式，其中 $m$ 和 $n$ 是正整数且 $n$ 没有被任何质数的平方整除。求 $m+n$？

(A) 65 65

(B) 132 132

(D) 194 194

(E) 215 215

Answer

Correct choice: (D)

正确答案：（D）

Solution

Angle chasing* reveals that $\triangle BPC\sim\triangle BDA$, therefore \[2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43},\] or $AB=86$. Additional angle chasing shows that $\triangle ABO\sim\triangle CDO$, therefore \[2=\frac{AB}{CD}=\frac{BO}{OD}=\frac{BP+11}{BP-11},\] or $BP=33$ and $BD=66$. Since $\triangle ADB$ is right, the Pythagorean theorem implies that \[AD=\sqrt{86^2-66^2}=4\sqrt{190}.\] The answer is $4+190=\boxed{\textbf{(D) }194}$. - Angle Chasing: If we set $\angle DBC = \alpha$, then we know that $\angle DCB = 180^\circ-2\alpha$ because $\triangle DBC$ is isosceles. Then, $\angle BCP = 90^\circ-\alpha$, so $\angle BPC$ is a right angle. Because $\angle BDC = \alpha$ and $\overline{AB}\parallel\overline{DC}$, we conclude that $\angle ABD = \alpha$ too. Lastly, because $\triangle BPC$ and $\triangle BDA$ are both right triangles, they are similar by AA.

通过角度追踪* 可知 $\triangle BPC\sim\triangle BDA$，因此 \[2=\frac{BD}{BP}=\frac{AB}{BC}=\frac{AB}{43},\] 即 $AB=86$。进一步的角度追踪显示 $\triangle ABO\sim\triangle CDO$，因此 \[2=\frac{AB}{CD}=\frac{BO}{OD}=\frac{BP+11}{BP-11},\] 即 $BP=33$，$BD=66$。由于 $\triangle ADB$ 是直角三角形，勾股定理得出 \[AD=\sqrt{86^2-66^2}=4\sqrt{190}.\] 答案是 $4+190=\boxed{\textbf{(D) }194}$。 - 角度追踪：设 $\angle DBC = \alpha$，则因为 $\triangle DBC$ 是等腰三角形，$\angle DCB = 180^\circ-2\alpha$。于是 $\angle BCP = 90^\circ-\alpha$，所以 $\angle BPC$ 是直角。因为 $\angle BDC = \alpha$ 且 $\overline{AB}\parallel\overline{DC}$，我们得出 $\angle ABD = \alpha$。最后，因为 $\triangle BPC$ 和 $\triangle BDA$ 都是直角三角形，它们通过 AA 相似。

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