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

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==Solution 3 (Nice, I Think?)==
 
==Solution 3 (Nice, I Think?)==
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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 answer is <math>80,</math> or <math>E.</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2018|ab=A|num-b=22|num-a=24}}
 
{{AMC12 box|year=2018|ab=A|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:16, 15 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?)

Consider the bisector of $\angle ATP.$ This angle makes an $80^{\circ}$ angle with $AP.$ We claim that $MN$ is parallel to this angle bisector, meaning that the answer is $80,$ or $E.$

See Also

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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