AMC12 2018 A

AMC12 2018 A · Q23

AMC12 2018 A · Q23. It mainly tests Angle chasing, Transformations.

In $\triangle PAT$, $\angle P = 36^\circ$, $\angle A = 56^\circ$, and $PA = 10$. Points $U$ and $G$ lie on sides $TP$ and $TA$, respectively, so that $PU = AG = 1$. Let $M$ and $N$ be the midpoints of segments $PA$ and $UG$, respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA$?

在 $\triangle PAT$ 中，$\angle P = 36^\circ$，$\angle A = 56^\circ$，$PA = 10$。点 $U$ 和 $G$ 分别在边 $TP$ 和 $TA$ 上，使得 $PU = AG = 1$。设 $M$ 和 $N$ 分别是线段 $PA$ 和 $UG$ 的中点。求直线 $MN$ 和 $PA$ 形成的锐角的度数。

(A) 76 76

(B) 77 77

(D) 79 79

(E) 80 80

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Extend $\overline{PN}$ through $N$ to $Q$ so that $PN = NQ$. Segments $\overline{UG}$ and $\overline{PQ}$ bisect each other, implying that $UPGQ$ is a parallelogram. Therefore $GQ \parallel PT$, so $\angle QGA = 180^\circ - \angle T = \angle TPA + \angle TAP = 36^\circ + 56^\circ = 92^\circ$. Furthermore $GQ = PU = AG$, so $\triangle QGA$ is isosceles, and $\angle QAG = \frac{1}{2}(180^\circ - 92^\circ) = 44^\circ$. Because $\overline{MN}$ is a midline of $\triangle QPA$, it follows that $\overline{MN} \parallel \overline{AQ}$ and \[ \angle NMP = \angle QAP = \angle QAG + \angle GAP = 44^\circ + 56^\circ = 100^\circ, \] so acute $\angle NMA = 80^\circ$. (Note that the value of the common length $PU = AG$ is immaterial.)

答案（E）：将$\overline{PN}$延长过$N$到点$Q$，使得$PN = NQ$。线段$\overline{UG}$与$\overline{PQ}$互相平分，因此$UPGQ$是平行四边形。于是$GQ \parallel PT$，所以$\angle QGA = 180^\circ - \angle T = \angle TPA + \angle TAP = 36^\circ + 56^\circ = 92^\circ$。又因为$GQ = PU = AG$，所以$\triangle QGA$是等腰三角形，从而$\angle QAG = \frac{1}{2}(180^\circ - 92^\circ) = 44^\circ$。由于$\overline{MN}$是$\triangle QPA$的一条中位线，可得$\overline{MN} \parallel \overline{AQ}$，并且 \[ \angle NMP = \angle QAP = \angle QAG + \angle GAP = 44^\circ + 56^\circ = 100^\circ, \] 因此锐角$\angle NMA = 80^\circ$。（注意公共长度$PU = AG$的具体取值并不重要。）

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