AMC12 2020 B

AMC12 2020 B · Q12

AMC12 2020 B · Q12. It mainly tests Circle theorems, Trigonometry (basic).

Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt{2}$. Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ so that $BE = 2\sqrt{5}$ and $\angle AEC = 45^\circ$. What is $CE^2 + DE^2$?

设$\overline{AB}$是一圆半径为$5\sqrt{2}$的直径。设$\overline{CD}$是圆内的一条弦，与$\overline{AB}$在点$E$相交，使得$BE = 2\sqrt{5}$且$\angle AEC = 45^\circ$。求$CE^2 + DE^2$？

(A) 96 96

(B) 98 98

(D) $70\sqrt{2}$ $70\sqrt{2}$

(E) 100 100

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $\overline{DF}$ be a chord of the circle that is perpendicular to the diameter $\overline{AB}$, and let $G$ be the intersection of $\overline{AB}$ and $\overline{DF}$. Let $O$ be the center of the circle. Because $\triangle EGD$ and $\triangle EGF$ are congruent and $\angle GED = 45^\circ$, it follows that $\triangle DEF$ is a right isosceles triangle with $EF = ED$. Then $\angle CDF = \angle EDF = 45^\circ$. Because $\angle COF$ subtends arc $CF$, it follows that $\angle COF = 2\angle CDF = 90^\circ$. Because $\triangle CEF$ and $\triangle COF$ are right triangles, $$ CE^2 + EF^2 = CF^2 = CO^2 + OF^2 = 2(5\sqrt{2})^2 = 100. $$ Thus $CE^2 + ED^2 = CE^2 + EF^2 = 100$. Notice that the position of $E$ on $\overline{AB}$ is irrelevant.

答案（E）：设$\overline{DF}$为圆的一条弦，且垂直于直径$\overline{AB}$，并令$G$为$\overline{AB}$与$\overline{DF}$的交点。设$O$为圆心。因为$\triangle EGD$与$\triangle EGF$全等且$\angle GED = 45^\circ$，可得$\triangle DEF$为等腰直角三角形，并且$EF = ED$。于是$\angle CDF = \angle EDF = 45^\circ$。由于$\angle COF$所对的是弧$CF$，可得$\angle COF = 2\angle CDF = 90^\circ$。因为$\triangle CEF$与$\triangle COF$都是直角三角形， $$ CE^2 + EF^2 = CF^2 = CO^2 + OF^2 = 2(5\sqrt{2})^2 = 100. $$ 因此$CE^2 + ED^2 = CE^2 + EF^2 = 100$。注意点$E$在$\overline{AB}$上的位置无关紧要。

Topics