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

(Solution 3 (Nice, I Think?))
(Solution 3 (Nice, I Think?))
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Let the bisector of <math>\angle ATP</math> intersect <math>PA</math> at <math>X.</math> We have <math>\angle ATX = \angle PTX = 44^{\circ},</math> so <math>\angle TXA = 80^{\circ}.</math> 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>.
 
Let the bisector of <math>\angle ATP</math> intersect <math>PA</math> at <math>X.</math> We have <math>\angle ATX = \angle PTX = 44^{\circ},</math> so <math>\angle TXA = 80^{\circ}.</math> 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>.
  
To prove this, let <math>N(x)</math> be the midpoint of <math>U(x)G(x),</math> where <math>U(x)</math> and <math>G(x)</math> are the points on <math>PT</math> and <math>AT,</math> respectively, such that <math>PU = AG = x.</math> (The points given in this problem correspond to <math>x=1,</math> but the idea we're getting at is that <math>x</math> will ultimately not matter.)
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To prove this, let <math>N(x)</math> be the midpoint of <math>U(x)G(x),</math> where <math>U(x)</math> and <math>G(x)</math> are the points on <math>PT</math> and <math>AT,</math> respectively, such that <math>PU = AG = x.</math> (The points given in this problem correspond to <math>x=1,</math> but the idea we're getting at is that <math>x</math> will ultimately not matter.) Since <math>U(x)</math> and <math>G(x)</math> vary linearly with <math>x,</math> the locus of all points <math>N(x)</math> must be a line. Let <math>Y</math> be the intersection of this line with <math>PT</math> (we know that this line will intersect <math>PT</math> and not <math>AT</math> because <math>PT > AT</math>).
  
The critical insight to finding this solution is that the length <math>1</math> probably shouldn't matter, because a length ratio of <math>1:5</math> or <math>1:10</math> (which show up in the problem) is exceedingly unlikely to generate nice angles. This idea then motivates the idea of looking at all points similar to <math>N,</math> which then leads to looking at the most convenient such point (in this case, the one that lies on <math>PT</math>).
+
The critical insight to finding this solution is that the length <math>1</math> probably shouldn't matter, because a length ratio of <math>1:5</math> or <math>1:10</math> (as in the problem) is exceedingly unlikely to generate nice angles. This idea then motivates the idea of looking at all points similar to <math>N,</math> which then leads to looking at the most convenient such point (in this case, the one that lies on <math>PT</math>).
  
 
(sujaykazi)
 
(sujaykazi)

Revision as of 00:33, 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?)

Let the bisector of $\angle ATP$ intersect $PA$ at $X.$ We have $\angle ATX = \angle PTX = 44^{\circ},$ so $\angle TXA = 80^{\circ}.$ 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 $\boxed{\textbf{(E)}}$.

To prove this, let $N(x)$ be the midpoint of $U(x)G(x),$ where $U(x)$ and $G(x)$ are the points on $PT$ and $AT,$ respectively, such that $PU = AG = x.$ (The points given in this problem correspond to $x=1,$ but the idea we're getting at is that $x$ will ultimately not matter.) Since $U(x)$ and $G(x)$ vary linearly with $x,$ the locus of all points $N(x)$ must be a line. Let $Y$ be the intersection of this line with $PT$ (we know that this line will intersect $PT$ and not $AT$ because $PT > AT$).

The critical insight to finding this solution is that the length $1$ probably shouldn't matter, because a length ratio of $1:5$ or $1:10$ (as in the problem) is exceedingly unlikely to generate nice angles. This idea then motivates the idea of looking at all points similar to $N,$ which then leads to looking at the most convenient such point (in this case, the one that lies on $PT$).

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

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
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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