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

Latest revision as of 23:53, 1 September 2024

1 Problem
2 Diagram
3 Solution 1 (Trigonometry)
4 Solution 2 (Rotation, Isosceles Triangle, Parallel Lines)
5 Solution 3 (Extending PN)
6 Solution 4 (Parallel Lines)
7 Solution 5 (Angle Bisectors)
8 Solution 6 (Overkill: Miquel Points)
9 Solution 7 (Olympiad Nuke)
10 Solution 8 (Vectors)
11 Video Solution by Richard Rusczyk
12 See Also

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 $\angle{DMN}+\angle{NMP}+\angle{P}=180^{\circ}$ , we can infer that $\angle{NMP}=100^{\circ}$ . 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 $AT = a, TP = b, PA = c.$ 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

@@ Line 140: / Line 140: @@
 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>.
+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{NMP}+\angle{P}=180^{\circ}</math>, we can infer that <math>\angle{NMP}=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>.

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
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