Difference between revisions of "2018 AMC 12A Problems/Problem 23"

Revision as of 23:20, 14 February 2018

Problem

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

$\textbf{(A)} 76 \qquad \textbf{(B)} 77 \qquad \textbf{(C)} 78 \qquad \textbf{(D)} 79 \qquad \textbf{(E)} 80$

Solution

Let $P$ be the origin, and $PA$ lie on the x axis.

We can find $U=\left(\cos(36), \sin(36)\right)$ and $G=\left(10-\cos(56), \sin(56)\right)$

Then, we have $M=(5, 0)$ and $N=\left(\frac{10+\cos(36)-\cos(56)}{2}, \frac{\sin(36)+\sin(56)}{2}\right)$

Notice that the tangent of our desired points is the the absolute difference between the y coordinates of the two points divided by the absolute difference between the x coordinates of the two points.

This evaluates to $\frac{\sin(36)+\sin(56)}{\cos(36)-\cos(56)}$ Now, using sum to product identities, we have this equal to $\frac{2\sin(46)\cos(10)}{-2\sin(46)\sin({-10})}=\frac{\sin(80)}{\cos(80)}=\tan(80)$ so the answer is $\boxed{\textbf{(E)}.}$ (lifeisgood03)

Solution 2 (Overkill)

Note that $X$ , the midpoint of major arc $PA$ on $(PAT)$ is the Miquel Point of $PUAG$ (Because $PU = AG$ ). Then, since $1 = \frac{UN}{NG} = \frac{PM}{MA}$ , this spiral similarity carries $M$ to $N$ . Thus, we have $\triangle XMN \sim \triangle XAG$ , so $\angle XMN = \angle XAG$ .

But, we have $\angle XAG = \angle PAG = \angle PAX = 56 - \frac{180 - \angle PXA}{2} =56 - \frac{180 - \angle T}{2} = 56 - \frac{\angle A + \angle P}{2} = 56 - \frac{56+36}{2} = 56 - 46 = 10$ ; thus $\angle XMN = 10$ .

Then, as $X$ is the midpoint of the major arc, it lies on the perpendicular bisector of $PA$ , so $\angle XMA = 90$ . Since we want the acute angle, we have $\angle NMA = \angle XMA - \angle XMN = 90 - 10 = 80$ , so the answer is $\boxed{\textbf{(E)}}$ .

(stronto)

Solution 3 (Nice, I Think?)

The bisector of $\angle ATP$ makes an $80^{\circ}$ angle with $PA$ by basic angle calculations. We claim that $MN$ is parallel to this angle bisector, meaning that the acute angle formed by $MN$ and $PA$ is $80^{\circ},$ meaning that the answer is $\boxed{\textbf{(E)}}$ .

(sujaykazi) Shoutout to Richard Yi and Mark Kong for working with me to discover the necessary insights to this problem!

@@ Line 35: / Line 35: @@
 ==Solution 3 (Nice, I Think?)==
-Consider the bisector of <math>\angle ATP.</math> This angle makes an <math>80^{\circ}</math> angle with <math>AP.</math> We claim that <math>MN</math> is parallel to this angle bisector, meaning that the acute angle that <math>MN</math> makes with <math>AP</math> is <math>80^{\circ},</math> meaning that the answer is <math>\boxed{\textbf{(E)}}</math>.
+The bisector of <math>\angle ATP</math> makes an <math>80^{\circ}</math> angle with <math>PA</math> by basic angle calculations. We claim that <math>MN</math> is parallel to this angle bisector, meaning that the acute angle formed by <math>MN</math> and <math>PA</math> is <math>80^{\circ},</math> meaning that the answer is <math>\boxed{\textbf{(E)}}</math>.
 (sujaykazi)
-Shoutout to Richard Yi and Mark Kong for working with me to glean the necessary insights for this problem!
+Shoutout to Richard Yi and Mark Kong for working with me to discover the necessary insights to this problem!
 ==See Also==
 {{AMC12 box|year=2018|ab=A|num-b=22|num-a=24}}
 {{MAA Notice}}

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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