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

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m (Solution 7 (Olympiad Nuke))
 
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== Problem ==
 
== Problem ==
 +
In <math>\triangle PAT,</math> <math>\angle P=36^{\circ},</math> <math>\angle A=56^{\circ},</math> and <math>PA=10.</math> Points <math>U</math> and <math>G</math> lie on sides <math>\overline{TP}</math> and <math>\overline{TA},</math> respectively, so that <math>PU=AG=1.</math> Let <math>M</math> and <math>N</math> be the midpoints of segments <math>\overline{PA}</math> and <math>\overline{UG},</math> respectively. What is the degree measure of the acute angle formed by lines <math>MN</math> and <math>PA?</math>
  
 +
<math>\textbf{(A) } 76 \qquad
 +
\textbf{(B) } 77 \qquad
 +
\textbf{(C) } 78 \qquad
 +
\textbf{(D) } 79 \qquad
 +
\textbf{(E) } 80 </math>
  
In <math>\triangle PAT,</math> <math>\angle P=36^{\circ},</math> <math>\angle A=56^{\circ},</math> and <math>PA=10.</math> Points <math>U</math> and <math>G</math> lie on sides <math>\overline{TP}</math> and <math>\overline{TA},</math> respectively, so that <math>PU=AG=1.</math> Let <math>M</math> and <math>N</math> be the midpoints of segments <math>\overline{PA}</math> and <math>\overline{UG},</math> respectively. What is the degree measure of the acute angle formed by lines <math>MN</math> and <math>PA?</math>
+
== Diagram ==
 
+
<asy>
<math>\textbf{(A)} 76 \qquad
+
/* Made by MRENTHUSIASM */
\textbf{(B)} 77 \qquad
+
size(375);
\textbf{(C)} 78 \qquad
 
\textbf{(D)} 79 \qquad
 
\textbf{(E)} 80 </math>
 
  
== Solution ==
+
pair P, A, T, U, G, M, N;
 +
P = origin;
 +
A = (10,0);
 +
U = intersectionpoint(Circle(P,1),P--P+2*dir(36));
 +
G = intersectionpoint(Circle(A,1),A--A+2*dir(180-56));
 +
T = extension(P,U,A,G);
 +
M = midpoint(P--A);
 +
N = midpoint(U--G);
 +
dot("$P$",P,1.5*SW,linewidth(4));
 +
dot("$A$",A,1.5*SE,linewidth(4));
 +
dot("$U$",U,1.5*(0,1),linewidth(4));
 +
dot("$G$",G,1.5*NE,linewidth(4));
 +
dot("$T$",T,1.5*(0,1),linewidth(4));
 +
dot("$M$",M,1.5*S,linewidth(4));
 +
dot("$N$",N,1.5*(0,1),linewidth(4));
 +
draw(P--A--T--cycle^^U--G^^M--N);
 +
label("$1$",midpoint(G--A),1.5*dir(30));
 +
label("$1$",midpoint(U--P),1.5*dir(150));
 +
label("$36^\circ$",P,5*dir(18),fontsize(10));
 +
label("$56^\circ$",A,3*dir(180-56/2),fontsize(10));
 +
Label L = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
 +
draw(P-(0,1)--A-(0,1), L=L, arrow=Arrows(),bar=Bars(15));
 +
add(pathticks(U--N, 2, .5, 4, 8, red));
 +
add(pathticks(N--G, 2, .5, 4, 8, red));
 +
add(pathticks(P--M, 1, .5, 0, 8, red));
 +
add(pathticks(M--A, 1, .5, 0, 8, red));
 +
</asy>
 +
~MRENTHUSIASM
  
Let <math>P</math> be the origin, and <math>PA</math> lie on the x axis.
+
== Solution 1 (Trigonometry) ==
 +
Let <math>P</math> be the origin, and <math>PA</math> lie on the <math>x</math>-axis.
  
 
We can find <math>U=\left(\cos(36), \sin(36)\right)</math> and <math>G=\left(10-\cos(56), \sin(56)\right)</math>
 
We can find <math>U=\left(\cos(36), \sin(36)\right)</math> and <math>G=\left(10-\cos(56), \sin(56)\right)</math>
Line 18: Line 49:
 
Then, we have <math>M=(5, 0)</math> and <math>N</math> is the midpoint of <math>U</math> and <math>G</math>, or <math>\left(\frac{10+\cos(36)-\cos(56)}{2}, \frac{\sin(36)+\sin(56)}{2}\right)</math>
 
Then, we have <math>M=(5, 0)</math> and <math>N</math> is the midpoint of <math>U</math> and <math>G</math>, or <math>\left(\frac{10+\cos(36)-\cos(56)}{2}, \frac{\sin(36)+\sin(56)}{2}\right)</math>
  
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.
+
Notice that the tangent of our desired points is the the absolute difference between the <math>y</math>-coordinates of the two points divided by the absolute difference between the <math>x</math>-coordinates of the two points.
  
 
This evaluates to <cmath>\frac{\sin(36)+\sin(56)}{\cos(36)-\cos(56)}</cmath>
 
This evaluates to <cmath>\frac{\sin(36)+\sin(56)}{\cos(36)-\cos(56)}</cmath>
 
Now, using sum to product identities, we have this equal to <cmath>\frac{2\sin(46)\cos(10)}{-2\sin(46)\sin({-10})}=\frac{\sin(80)}{\cos(80)}=\tan(80)</cmath>
 
Now, using sum to product identities, we have this equal to <cmath>\frac{2\sin(46)\cos(10)}{-2\sin(46)\sin({-10})}=\frac{\sin(80)}{\cos(80)}=\tan(80)</cmath>
so the answer is <math>\boxed{\textbf{(E)}.}</math> (lifeisgood03)
+
so the answer is <math>\boxed{\textbf{(E) } 80}.</math>
 +
 
 +
~lifeisgood03
 +
 
 +
Note: Though this solution is excellent, setting <math>M = (0,0)</math> makes life a tad bit easier
 +
 
 +
~MathleteMA
 +
 
 +
== Solution 2 (Rotation, Isosceles Triangle, Parallel Lines) ==
 +
We will refer to the <b>Diagram</b> section. In this solution, 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.
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(375);
 +
 
 +
pair P, A, T, U, G, M, N, U1, H;
 +
P = origin;
 +
A = (10,0);
 +
U = intersectionpoint(Circle(P,1),P--P+2*dir(36));
 +
G = intersectionpoint(Circle(A,1),A--A+2*dir(180-56));
 +
T = extension(P,U,A,G);
 +
M = midpoint(P--A);
 +
N = midpoint(U--G);
 +
U1 = rotate(180,M)*U;
 +
H = intersectionpoint(P--A,G--U1);
 +
fill(U--P--M--cycle^^M--U1--A--cycle,yellow);
 +
dot("$P$",P,1.5*SW,linewidth(4));
 +
dot("$A$",A,1.5*SE,linewidth(4));
 +
dot("$U$",U,1.5*(0,1),linewidth(4));
 +
dot("$G$",G,1.5*NE,linewidth(4));
 +
dot("$T$",T,1.5*(0,1),linewidth(4));
 +
dot("$M$",M,1.5*S,linewidth(4));
 +
dot("$N$",N,1.5*(0,1),linewidth(4));
 +
dot("$U'$",U1,1.5*S,linewidth(4));
 +
dot("$H$",H,1.5*NW,linewidth(4));
 +
draw(P--A--T--cycle^^U--G^^M--N^^U--U1--A);
 +
draw(G--U1,dashed);
 +
label("$1$",midpoint(G--A),1.5*dir(30));
 +
label("$1$",midpoint(A--U1),1.5*dir(-30));
 +
label("$1$",midpoint(U--P),1.5*dir(150));
 +
label("$36^\circ$",P,5*dir(18),fontsize(8));
 +
label("$56^\circ$",A,2.5*dir(180-56/2),fontsize(8));
 +
label("$36^\circ$",A,2.5*dir(180+25),fontsize(8));
 +
Label L = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
 +
draw(P-(0,1.5)--A-(0,1.5), L=L, arrow=Arrows(),bar=Bars(15));
 +
add(pathticks(U--N, 2, .5, 4, 8, red));
 +
add(pathticks(N--G, 2, .5, 4, 8, red));
 +
add(pathticks(U--M, 1, .5, 0, 8, red));
 +
add(pathticks(M--U1, 1, .5, 0, 8, red));
 +
</asy>
 +
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>
  
Note: Though this solution is excellent, setting <math>M = (0,0)</math> makes life a tad bit easier ~ MathleteMA
+
Finally, we have <math>\angle NMA=\angle GHA=\boxed{\textbf{(E) } 80}</math> by the Corresponding Angles Postulate.
  
==Solution 2 (Overkill)==
+
~MRENTHUSIASM
  
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 3 (Extending PN)==
 +
 
 +
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>.
 +
 
 +
<math>\because M,N</math> are the midpoints of <math>PA</math> and <math>PQ,</math> respectively
 +
 
 +
<math>\therefore MN</math> is the midsegment of <math>\bigtriangleup PAQ</math>
 +
 
 +
<math>\therefore \angle QAP=\angle NMP</math>
 +
 
 +
Notice that <math>\bigtriangleup PUN\cong \bigtriangleup QGN</math>
 +
 
 +
As a result, <math>QG=AG=UP=1</math>, <math>\angle AQG=\angle QAG</math>, <math>\angle GQN=\angle NPU</math>
 +
 
 +
Also, <math>\angle GQN+\angle QPA=\angle QPU+\angle QPA=\angle UPA=36^{\circ}</math>
 +
 
 +
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>
 +
 
 +
Since we are asked for the acute angle between the two lines, the answer to this problem is <math>\boxed{\textbf{(E) } 80}</math>.
  
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 by <math>BladeRunnerAUG</math> (Frank FYC)
  
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)}}</math>.
+
==Solution 4 (Parallel Lines)==
 +
Let the mid-point of <math>\overline{AT}</math> be <math>B</math> and the mid-point of <math>\overline{GT}</math> be <math>C</math>.
 +
Since <math>BC=CG-BG</math> and <math>CG=AB-\frac{1}{2}</math>, we can conclude that <math>BC=\frac{1}{2}</math>.
 +
Similarly, we can conclude that <math>BM-CN=\frac{1}{2}</math>. Construct <math>\overline{ND}\parallel\overline{BC}</math> and intersects <math>\overline{BM}</math> at <math>D</math>, which gives <math>MD=DN=\frac{1}{2}</math>.  
 +
Since <math>\angle{ABD}=\angle{BDN}</math>, <math>MD=DN</math>, we can find the value of <math>\angle{DMN}</math>, which is equal to <math>\frac{1}{2}\angle T=44^{\circ}</math>. Since <math>\overline{BM}\parallel\overline{PT}</math>, which means <math>\angle{DMN}+\angle{MNP}+\angle{P}=180^{\circ}</math>, we can infer that <math>\angle{MNP}=100^{\circ}</math>.
 +
As we are required to give the acute angle formed, the final answer would be <math>80^{\circ}</math>, which is <math>\boxed{\textbf{(E) } 80}</math>.
  
(stronto)
+
~Surefire2019
  
==Solution 3 (Nice, I Think?)==
+
==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)}}</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>
 
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>
Line 50: Line 159:
 
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>).
 
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>).
  
(sujaykazi)
+
~sujaykazi
 +
 
 
Shoutout to Richard Yi and Mark Kong for working with me to discover the necessary insights to this problem!
 
Shoutout to Richard Yi and Mark Kong for working with me to discover the necessary insights to this problem!
  
==Solution 4==
+
==Solution 6 (Overkill: Miquel Points)==
Let the mid-point of <math>\overline{AT}</math> be <math>B</math> and the mid-point of <math>\overline{GT}</math> be <math>C</math>.
 
Since <math>\overline{BC}=\overline{CG}-\overline{BG}</math> and <math>\overline{CG}=\overline{AB}-\frac{1}{2}</math>, we can conclude that <math>\overline{BC}=\frac{1}{2}</math>.
 
Similarly, we can conclude that <math>\overline{BM}-\overline{CN}=\frac{1}{2}</math>. Construct <math>ND//BC</math> and intersects <math>\overline{BM}</math> at <math>D</math>, which gives <math>\overline{MD}=\overline{DN}=\frac{1}{2}</math>.
 
Since <math>\angle{ABD}=\angle{BDN}</math>, <math>\overline{MD}=\overline{DN}</math>, we can find the value of <math>\angle{DMN}</math>, which is equal to <math>\frac{1}{2}T=44^{\circ}</math>. Since <math>BM//PT</math>, which means <math>\angle{DMN}+\angle{MNP}+\angle{P}=180^{\circ}</math>, we can infer that <math>\angle{MNP}=100^{\circ}</math>.
 
As we are required to give the acute angle formed, the final answer would be <math>80^{\circ}</math>, which is <math>\boxed{\textbf{(E)}}</math>.
 
(Surefire2019)
 
  
==Solution 5 (Simplest, I think)==
+
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>.
  
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>.  
+
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>.  
  
<math>\because M</math>, <math>N</math> is the middle point of <math>AP</math> and <math>QU</math>
+
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>.
  
<math>\therefore MN</math> is the middle line of <math>\bigtriangleup PAQ</math>
+
~stronto
  
<math>\therefore \angle QAP=\angle NMP</math>
+
<u><b>Sidenote</b></u>
  
Notice that <math>\bigtriangleup PUN\cong \bigtriangleup QGN</math>
+
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.
  
As a result, <math>QG=AG=UP=1</math>, <math>\angle AQG=\angle QAG</math>, <math>\angle GQN=\angle NPU</math>
+
== 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, <cmath>\angle NMA = 180^\circ - 56^\circ - \frac{180^\circ - 56^\circ - 36^\circ}{2} = \boxed{\textbf{(E) } 80}.</cmath>
  
Also, <math>\angle GQN+\angle QPA=\angle QPU+\angle QPA=\angle UPA=36^{\circ}</math>
+
== Solution 8 (Vectors) ==
  
As a result, <math>2\angle QAG=180^{\circ}-56^{\circ}-36^{\circ}=88^{\circ}</math>
+
The argument of the average of any two unit vectors is average of the arguments of the two vectors. Thereby, the acute angle formed is <cmath>\frac{36^\circ{} + 180^\circ{} - 56^\circ{}}{2} = \boxed{\textbf{(E) } 80}.</cmath>
  
Therefore, <math>\angle QAP=\angle QAG+\angle TAP=56^{\circ}+44^{\circ}=100^{\circ}</math>
+
~Professor-Mom (all credit for this amazing solution goes to V_Enhance)
 
 
Since we are asked for the acute angle between the two lines, the answer to this problem is <math>\boxed{80^{\circ}}</math>
 
 
 
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)
 
  
=== Video Solution by Richard Rusczyk ===
+
== Video Solution by Richard Rusczyk ==
  
 
https://artofproblemsolving.com/videos/amc/2018amc12a/473
 
https://artofproblemsolving.com/videos/amc/2018amc12a/473

Latest revision as of 22:11, 23 October 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

[asy] /* Made by MRENTHUSIASM */ size(375);  pair P, A, T, U, G, M, N; P = origin; A = (10,0); U = intersectionpoint(Circle(P,1),P--P+2*dir(36)); G = intersectionpoint(Circle(A,1),A--A+2*dir(180-56)); T = extension(P,U,A,G); M = midpoint(P--A); N = midpoint(U--G); dot("$P$",P,1.5*SW,linewidth(4)); dot("$A$",A,1.5*SE,linewidth(4)); dot("$U$",U,1.5*(0,1),linewidth(4)); dot("$G$",G,1.5*NE,linewidth(4)); dot("$T$",T,1.5*(0,1),linewidth(4)); dot("$M$",M,1.5*S,linewidth(4)); dot("$N$",N,1.5*(0,1),linewidth(4)); draw(P--A--T--cycle^^U--G^^M--N); label("$1$",midpoint(G--A),1.5*dir(30)); label("$1$",midpoint(U--P),1.5*dir(150)); label("$36^\circ$",P,5*dir(18),fontsize(10)); label("$56^\circ$",A,3*dir(180-56/2),fontsize(10)); Label L = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(P-(0,1)--A-(0,1), L=L, arrow=Arrows(),bar=Bars(15)); add(pathticks(U--N, 2, .5, 4, 8, red)); add(pathticks(N--G, 2, .5, 4, 8, red)); add(pathticks(P--M, 1, .5, 0, 8, red)); add(pathticks(M--A, 1, .5, 0, 8, red)); [/asy] ~MRENTHUSIASM

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. [asy] /* Made by MRENTHUSIASM */ size(375);  pair P, A, T, U, G, M, N, U1, H; P = origin; A = (10,0); U = intersectionpoint(Circle(P,1),P--P+2*dir(36)); G = intersectionpoint(Circle(A,1),A--A+2*dir(180-56)); T = extension(P,U,A,G); M = midpoint(P--A); N = midpoint(U--G); U1 = rotate(180,M)*U; H = intersectionpoint(P--A,G--U1); fill(U--P--M--cycle^^M--U1--A--cycle,yellow); dot("$P$",P,1.5*SW,linewidth(4)); dot("$A$",A,1.5*SE,linewidth(4)); dot("$U$",U,1.5*(0,1),linewidth(4)); dot("$G$",G,1.5*NE,linewidth(4)); dot("$T$",T,1.5*(0,1),linewidth(4)); dot("$M$",M,1.5*S,linewidth(4)); dot("$N$",N,1.5*(0,1),linewidth(4)); dot("$U'$",U1,1.5*S,linewidth(4)); dot("$H$",H,1.5*NW,linewidth(4)); draw(P--A--T--cycle^^U--G^^M--N^^U--U1--A); draw(G--U1,dashed); label("$1$",midpoint(G--A),1.5*dir(30)); label("$1$",midpoint(A--U1),1.5*dir(-30)); label("$1$",midpoint(U--P),1.5*dir(150)); label("$36^\circ$",P,5*dir(18),fontsize(8)); label("$56^\circ$",A,2.5*dir(180-56/2),fontsize(8)); label("$36^\circ$",A,2.5*dir(180+25),fontsize(8)); Label L = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(P-(0,1.5)--A-(0,1.5), L=L, arrow=Arrows(),bar=Bars(15)); add(pathticks(U--N, 2, .5, 4, 8, red)); add(pathticks(N--G, 2, .5, 4, 8, red)); add(pathticks(U--M, 1, .5, 0, 8, red)); add(pathticks(M--U1, 1, .5, 0, 8, red)); [/asy] 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 (Parallel Lines)

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, \[\angle NMA = 180^\circ - 56^\circ - \frac{180^\circ - 56^\circ - 36^\circ}{2} = \boxed{\textbf{(E) } 80}.\]

Solution 8 (Vectors)

The argument of the average of any two unit vectors is average of the arguments of the two vectors. Thereby, the acute angle formed is \[\frac{36^\circ{} + 180^\circ{} - 56^\circ{}}{2} = \boxed{\textbf{(E) } 80}.\]

~Professor-Mom (all credit for this amazing solution goes to V_Enhance)

Video Solution by Richard Rusczyk

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

~ dolphin7

See Also

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