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

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~MathleteMA
 
~MathleteMA
  
==Solution 2 (Overkill: Miquel Points)==
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== Solution 2 (Rotation, Isosceles Triangle, Parallel Lines) ==
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We will refer to the <b>Diagram</b> section. In this solution, all angle measures are in degrees.
  
Note that <math>X</math>, the midpoint of major arc <math>PA</math> on <math>(PAT)</math> is the Miquel Point of <math>PUAG</math> (Because <math>PU = AG</math>). Then, since <math>1 = \frac{UN}{NG} = \frac{PM}{MA}</math>, this spiral similarity carries <math>M</math> to <math>N</math>. Thus, we have <math>\triangle XMN \sim \triangle XAG</math>, so <math>\angle XMN = \angle XAG</math>.  
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We rotate <math>\triangle PUM</math> by <math>180^\circ</math> about <math>M</math> to obtain <math>\triangle AU'M.</math> Let <math>H</math> be the intersection of <math>\overline{PA}</math> and <math>\overline{GU'},</math> as shown below.
 +
[[File:2018 AMC 12A Problem 23 Solution.png|center|600px]]
 +
Note that <math>\triangle GU'A</math> is an isosceles triangle with <math>GA=U'A=1,</math> so <math>\angle AGU'=\angle AU'G=\frac{180-\angle GAU'}{2}=44.</math> In <math>\triangle GHA,</math> it follows that <math>\angle GHA=180-\angle GAH-\angle AGH=80.</math>
 +
 
 +
Since <math>\frac{UM}{UU'}=\frac{UN}{UG}=\frac12,</math> we conclude that <math>\triangle UMN\sim\triangle UU'G</math> by SAS, from which <math>\angle UMN=\angle UU'G</math> and <math>\angle UNM=\angle UGU'.</math> By the Converse of the Corresponding Angles Postulate, we deduce that <math>\overline{MN}\parallel\overline{U'G}.</math>
 +
 
 +
Finally, we have <math>\angle NMA=\angle GHA=\boxed{\textbf{(E) } 80}</math> by the Corresponding Angles Postulate.
  
But, we have <math>\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</math>; thus <math>\angle XMN = 10</math>.
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~MRENTHUSIASM
  
Then, as <math>X</math> is the midpoint of the major arc, it lies on the perpendicular bisector of <math>PA</math>, so <math>\angle XMA = 90</math>. Since we want the acute angle, we have <math>\angle NMA = \angle XMA - \angle XMN = 90 - 10 = 80</math>, so the answer is <math>\boxed{\textbf{(E) } 80}</math>.
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==Solution 3 (Extending PN)==
  
~stronto
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Link <math>PN</math>, extend <math>PN</math> to <math>Q</math> so that <math>QN=PN</math>. Then link <math>QG</math> and <math>QA</math>.
  
<u><b>Sidenote</b></u>
+
<math>\because M,N</math> are the midpoints of <math>PA</math> and <math>PQ,</math> respectively
  
For another way to find <math>\angle XMN</math>, note that <cmath>\angle XAM = 90 - \angle MXA = 90 - \frac{\angle AXP}{2} = 90 - \frac{\angle ATP}{2}= 90 - 44 = 46,</cmath> giving <math>\angle XMN = \angle XAG = 56 - 46 = 10</math> as desired.
+
<math>\therefore MN</math> is the midsegment of <math>\bigtriangleup PAQ</math>
  
==Solution 3 (Angle Bisectors)==
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<math>\therefore \angle QAP=\angle NMP</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) } 80}</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.) 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. Notice that <math>N(0) = M,</math> so <math>M</math> lies on this line. Let <math>N(x_0)</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>). Notice that <math>G(x_0) = T.</math>
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Notice that <math>\bigtriangleup PUN\cong \bigtriangleup QGN</math>
  
Let <math>AT = a, TP = b, PT = c.</math> Then <math>AG(x_0) = PU(x_0) = AT = a</math> and <math>PG(x_0) = PT = b.</math> Thus, <math>PN(x_0) = \frac{a+b}{2}.</math> By the Angle Bisector Theorem, <math>\frac{PX}{AX} = \frac{PT}{AT} = \frac{b}{a},</math> so <math>PX = \frac{bc}{a+b}.</math> Since <math>M</math> is the midpoint of <math>AP,</math> we also have <math>PM = \frac{c}{2}.</math> Notice that:
+
As a result, <math>QG=AG=UP=1</math>, <math>\angle AQG=\angle QAG</math>, <math>\angle GQN=\angle NPU</math>
  
<cmath>\frac{PM}{PX} = \frac{\frac{c}{2}}{\frac{bc}{a+b}} = \frac{a+b}{2b}</cmath>
+
Also, <math>\angle GQN+\angle QPA=\angle QPU+\angle QPA=\angle UPA=36^{\circ}</math>
<cmath>\frac{PN(x_0)}{PT} = \frac{\frac{a+b}{2}}{b} = \frac{a+b}{2b}</cmath>
 
  
Since <math>\frac{PN(x_0)}{PT} = \frac{PM}{PX},</math> the line containing all points <math>N(x)</math> must be parallel to <math>TX.</math> This concludes the proof.
+
As a result, <math>2\angle QAG=180^{\circ}-56^{\circ}-36^{\circ}=88^{\circ}</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 realization 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>).
+
Therefore, <math>\angle QAP=\angle QAG+\angle TAP=56^{\circ}+44^{\circ}=100^{\circ}</math>
  
~sujaykazi
+
Since we are asked for the acute angle between the two lines, the answer to this problem is <math>\boxed{\textbf{(E) } 80}</math>.
  
Shoutout to Richard Yi and Mark Kong for working with me to discover the necessary insights to this problem!
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~Solution by <math>BladeRunnerAUG</math> (Frank FYC)
  
 
==Solution 4==
 
==Solution 4==
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~Surefire2019
 
~Surefire2019
  
==Solution 5 (Extending PN)==
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==Solution 5 (Angle Bisectors)==
 +
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) } 80}</math>.
  
Link <math>PN</math>, extend <math>PN</math> to <math>Q</math> so that <math>QN=PN</math>. Then link <math>QG</math> and <math>QA</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.) 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. Notice that <math>N(0) = M,</math> so <math>M</math> lies on this line. Let <math>N(x_0)</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>). Notice that <math>G(x_0) = T.</math>
  
<math>\because M,N</math> are the midpoints of <math>PA</math> and <math>PQ,</math> respectively
+
Let <math>AT = a, TP = b, PT = c.</math> Then <math>AG(x_0) = PU(x_0) = AT = a</math> and <math>PG(x_0) = PT = b.</math> Thus, <math>PN(x_0) = \frac{a+b}{2}.</math> By the Angle Bisector Theorem, <math>\frac{PX}{AX} = \frac{PT}{AT} = \frac{b}{a},</math> so <math>PX = \frac{bc}{a+b}.</math> Since <math>M</math> is the midpoint of <math>AP,</math> we also have <math>PM = \frac{c}{2}.</math> Notice that:
  
<math>\therefore MN</math> is the midsegment of <math>\bigtriangleup PAQ</math>
+
<cmath>\frac{PM}{PX} = \frac{\frac{c}{2}}{\frac{bc}{a+b}} = \frac{a+b}{2b}</cmath>
 +
<cmath>\frac{PN(x_0)}{PT} = \frac{\frac{a+b}{2}}{b} = \frac{a+b}{2b}</cmath>
  
<math>\therefore \angle QAP=\angle NMP</math>
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Since <math>\frac{PN(x_0)}{PT} = \frac{PM}{PX},</math> the line containing all points <math>N(x)</math> must be parallel to <math>TX.</math> This concludes the proof.
  
Notice that <math>\bigtriangleup PUN\cong \bigtriangleup QGN</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 realization 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>).
  
As a result, <math>QG=AG=UP=1</math>, <math>\angle AQG=\angle QAG</math>, <math>\angle GQN=\angle NPU</math>
+
~sujaykazi
  
Also, <math>\angle GQN+\angle QPA=\angle QPU+\angle QPA=\angle UPA=36^{\circ}</math>
+
Shoutout to Richard Yi and Mark Kong for working with me to discover the necessary insights to this problem!
 
 
As a result, <math>2\angle QAG=180^{\circ}-56^{\circ}-36^{\circ}=88^{\circ}</math>
 
  
Therefore, <math>\angle QAP=\angle QAG+\angle TAP=56^{\circ}+44^{\circ}=100^{\circ}</math>
+
==Solution 6 (Overkill: Miquel Points)==
  
Since we are asked for the acute angle between the two lines, the answer to this problem is <math>\boxed{\textbf{(E) } 80}</math>.
+
Note that <math>X</math>, the midpoint of major arc <math>PA</math> on <math>(PAT)</math> is the Miquel Point of <math>PUAG</math> (Because <math>PU = AG</math>). Then, since <math>1 = \frac{UN}{NG} = \frac{PM}{MA}</math>, this spiral similarity carries <math>M</math> to <math>N</math>. Thus, we have <math>\triangle XMN \sim \triangle XAG</math>, so <math>\angle XMN = \angle XAG</math>.  
  
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)
+
But, we have <math>\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</math>; thus <math>\angle XMN = 10</math>.
  
== Solution 6 (Olympiad Nuke) ==
+
Then, as <math>X</math> is the midpoint of the major arc, it lies on the perpendicular bisector of <math>PA</math>, so <math>\angle XMA = 90</math>. Since we want the acute angle, we have <math>\angle NMA = \angle XMA - \angle XMN = 90 - 10 = 80</math>, so the answer is <math>\boxed{\textbf{(E) } 80}</math>.
By https://artofproblemsolving.com/community/c6h489748p2745891, we get that <math>MN</math> is parallel to the angle bisector of <math>\angle ATP.</math> Thus, <math>\angle MNA = 180^\circ - 56^\circ - (180^\circ - 56^\circ - 36^\circ)/2 = 80^\circ \implies \boxed{\textbf{(E) } 80}.</math>
 
  
== Solution 7 (Rotation, Isosceles Triangle, Parallel Lines) ==
+
~stronto
This solution refers to the <b>Diagram</b> section. Furthermore, all angle measures are in degrees.
 
  
We rotate <math>\triangle PUM</math> by <math>180^\circ</math> about <math>M</math> to obtain <math>\triangle AU'M.</math> Let <math>H</math> be the intersection of <math>\overline{PA}</math> and <math>\overline{GU'},</math> as shown below.
+
<u><b>Sidenote</b></u>
[[File:2018 AMC 12A Problem 23 Solution.png|center|600px]]
 
Note that <math>\triangle GU'A</math> is an isosceles triangle with <math>GA=U'A=1,</math> so <math>\angle AGU'=\angle AU'G=\frac{180-\angle GAU'}{2}=44.</math> In <math>\triangle GHA,</math> it follows that <math>\angle GHA=180-\angle GAH-\angle AGH=80.</math>
 
  
Since <math>\frac{UM}{UU'}=\frac{UN}{UG}=\frac12,</math> we conclude that <math>\triangle UMN\sim\triangle UU'G</math> by SAS, from which <math>\angle UMN=\angle UU'G</math> and <math>\angle UNM=\angle UGU'.</math> By the Converse of the Corresponding Angles Postulate, we deduce that <math>\overline{MN}\parallel\overline{U'G}.</math> Finally, we have <math>\angle NMA=\angle GHA=\boxed{\textbf{(E) } 80}</math> by the Corresponding Angles Postulate.
+
For another way to find <math>\angle XMN</math>, note that <cmath>\angle XAM = 90 - \angle MXA = 90 - \frac{\angle AXP}{2} = 90 - \frac{\angle ATP}{2}= 90 - 44 = 46,</cmath> giving <math>\angle XMN = \angle XAG = 56 - 46 = 10</math> as desired.
  
~MRENTHUSIASM
+
== Solution 7 (Olympiad Nuke) ==
 +
By https://artofproblemsolving.com/community/c6h489748p2745891, we get that <math>MN</math> is parallel to the angle bisector of <math>\angle ATP.</math> Thus, <math>\angle MNA = 180^\circ - 56^\circ - (180^\circ - 56^\circ - 36^\circ)/2 = 80^\circ \implies \boxed{\textbf{(E) } 80}.</math>
  
 
== Video Solution by Richard Rusczyk ==
 
== Video Solution by Richard Rusczyk ==

Revision as of 22:51, 23 June 2021

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 $PA?$

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

Diagram

~MRENTHUSIASM (by Geometry Expressions)

Solution 1 (Trigonometry)

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 $N$ is the midpoint of $U$ and $G$, or $\left(\frac{10+\cos(36)-\cos(56)}{2}, \frac{\sin(36)+\sin(56)}{2}\right)$

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 $\boxed{\textbf{(E) } 80}.$

~lifeisgood03

Note: Though this solution is excellent, setting $M = (0,0)$ makes life a tad bit easier

~MathleteMA

Solution 2 (Rotation, Isosceles Triangle, Parallel Lines)

We will refer to the Diagram section. In this solution, all angle measures are in degrees.

We rotate $\triangle PUM$ by $180^\circ$ about $M$ to obtain $\triangle AU'M.$ Let $H$ be the intersection of $\overline{PA}$ and $\overline{GU'},$ as shown below.

Note that $\triangle GU'A$ is an isosceles triangle with $GA=U'A=1,$ so $\angle AGU'=\angle AU'G=\frac{180-\angle GAU'}{2}=44.$ In $\triangle GHA,$ it follows that $\angle GHA=180-\angle GAH-\angle AGH=80.$

Since $\frac{UM}{UU'}=\frac{UN}{UG}=\frac12,$ we conclude that $\triangle UMN\sim\triangle UU'G$ by SAS, from which $\angle UMN=\angle UU'G$ and $\angle UNM=\angle UGU'.$ By the Converse of the Corresponding Angles Postulate, we deduce that $\overline{MN}\parallel\overline{U'G}.$

Finally, we have $\angle NMA=\angle GHA=\boxed{\textbf{(E) } 80}$ by the Corresponding Angles Postulate.

~MRENTHUSIASM

Solution 3 (Extending PN)

Link $PN$, extend $PN$ to $Q$ so that $QN=PN$. Then link $QG$ and $QA$.

$\because M,N$ are the midpoints of $PA$ and $PQ,$ respectively

$\therefore MN$ is the midsegment of $\bigtriangleup PAQ$

$\therefore \angle QAP=\angle NMP$

Notice that $\bigtriangleup PUN\cong \bigtriangleup QGN$

As a result, $QG=AG=UP=1$, $\angle AQG=\angle QAG$, $\angle GQN=\angle NPU$

Also, $\angle GQN+\angle QPA=\angle QPU+\angle QPA=\angle UPA=36^{\circ}$

As a result, $2\angle QAG=180^{\circ}-56^{\circ}-36^{\circ}=88^{\circ}$

Therefore, $\angle QAP=\angle QAG+\angle TAP=56^{\circ}+44^{\circ}=100^{\circ}$

Since we are asked for the acute angle between the two lines, the answer to this problem is $\boxed{\textbf{(E) } 80}$.

~Solution by $BladeRunnerAUG$ (Frank FYC)

Solution 4

Let the mid-point of $\overline{AT}$ be $B$ and the mid-point of $\overline{GT}$ be $C$. Since $BC=CG-BG$ and $CG=AB-\frac{1}{2}$, we can conclude that $BC=\frac{1}{2}$. Similarly, we can conclude that $BM-CN=\frac{1}{2}$. Construct $\overline{ND}\parallel\overline{BC}$ and intersects $\overline{BM}$ at $D$, which gives $MD=DN=\frac{1}{2}$. Since $\angle{ABD}=\angle{BDN}$, $MD=DN$, we can find the value of $\angle{DMN}$, which is equal to $\frac{1}{2}\angle T=44^{\circ}$. Since $\overline{BM}\parallel\overline{PT}$, which means $\angle{DMN}+\angle{MNP}+\angle{P}=180^{\circ}$, we can infer that $\angle{MNP}=100^{\circ}$. As we are required to give the acute angle formed, the final answer would be $80^{\circ}$, which is $\boxed{\textbf{(E) } 80}$.

~Surefire2019

Solution 5 (Angle Bisectors)

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 $\boxed{\textbf{(E) } 80}$.

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. Notice that $N(0) = M,$ so $M$ lies on this line. Let $N(x_0)$ be the intersection of this line with $PT$ (we know that this line will intersect $PT$ and not $AT$ because $PT > AT$). Notice that $G(x_0) = T.$

Let $AT = a, TP = b, PT = c.$ Then $AG(x_0) = PU(x_0) = AT = a$ and $PG(x_0) = PT = b.$ Thus, $PN(x_0) = \frac{a+b}{2}.$ By the Angle Bisector Theorem, $\frac{PX}{AX} = \frac{PT}{AT} = \frac{b}{a},$ so $PX = \frac{bc}{a+b}.$ Since $M$ is the midpoint of $AP,$ we also have $PM = \frac{c}{2}.$ Notice that:

\[\frac{PM}{PX} = \frac{\frac{c}{2}}{\frac{bc}{a+b}} = \frac{a+b}{2b}\] \[\frac{PN(x_0)}{PT} = \frac{\frac{a+b}{2}}{b} = \frac{a+b}{2b}\]

Since $\frac{PN(x_0)}{PT} = \frac{PM}{PX},$ the line containing all points $N(x)$ must be parallel to $TX.$ This concludes the proof.

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 realization 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!

Solution 6 (Overkill: Miquel Points)

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 $\boxed{\textbf{(E) } 80}$.

~stronto

Sidenote

For another way to find $\angle XMN$, note that \[\angle XAM = 90 - \angle MXA = 90 - \frac{\angle AXP}{2} = 90 - \frac{\angle ATP}{2}= 90 - 44 = 46,\] giving $\angle XMN = \angle XAG = 56 - 46 = 10$ as desired.

Solution 7 (Olympiad Nuke)

By https://artofproblemsolving.com/community/c6h489748p2745891, we get that $MN$ is parallel to the angle bisector of $\angle ATP.$ Thus, $\angle MNA = 180^\circ - 56^\circ - (180^\circ - 56^\circ - 36^\circ)/2 = 80^\circ \implies \boxed{\textbf{(E) } 80}.$

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2018amc12a/473

~ dolphin7

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
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Problem 22
Followed by
Problem 24
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