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

m (→‎Solution 4: // -> \parallel)
(Reformatted the titles so that they look more professional. Also, I fixed some of the logical errors. Let me know if anyone is unhappy with this.)
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~MRENTHUSIASM (by Geometry Expressions)
 
~MRENTHUSIASM (by Geometry Expressions)
  
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== Solution 1 ==
+
== Solution 1 (Trigonometry) ==
 
Let <math>P</math> be the origin, and <math>PA</math> lie on the <math>x</math>-axis.
 
Let <math>P</math> be the origin, and <math>PA</math> lie on the <math>x</math>-axis.
  
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~MathleteMA
 
~MathleteMA
  
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==Solution 2 (Overkill)==
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==Solution 2 (Overkill: Miquel Points)==
  
 
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>.  
 
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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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.
 
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.
  
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==Solution 3 (Nice, I Think?)==
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==Solution 3 (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>.
 
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>.
  
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~Surefire2019
 
~Surefire2019
  
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==Solution 5 (Simplest, I think?)==
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==Solution 5 (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>.  
 
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</math>, <math>N</math> is the middle point of <math>AP</math> and <math>QU</math>
+
<math>\because M,N</math> are the midpoints of <math>PA</math> and <math>PQ,</math> respectively
  
−
<math>\therefore MN</math> is the middle line of <math>\bigtriangleup PAQ</math>
+
<math>\therefore MN</math> is the midsegment of <math>\bigtriangleup PAQ</math>
  
 
<math>\therefore \angle QAP=\angle NMP</math>
 
<math>\therefore \angle QAP=\angle NMP</math>

Revision as of 23:22, 22 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

600px

~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 (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 3 (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 4

Let the mid-point of $\overline{AT}$ be $B$ and the mid-point of $\overline{GT}$ be $C$. Since $\overline{BC}=\overline{CG}-\overline{BG}$ and $\overline{CG}=\overline{AB}-\frac{1}{2}$, we can conclude that $\overline{BC}=\frac{1}{2}$. Similarly, we can conclude that $\overline{BM}-\overline{CN}=\frac{1}{2}$. Construct $ND\parallel BC$ and intersects $\overline{BM}$ at $D$, which gives $\overline{MD}=\overline{DN}=\frac{1}{2}$. Since $\angle{ABD}=\angle{BDN}$, $\overline{MD}=\overline{DN}$, we can find the value of $\angle{DMN}$, which is equal to $\frac{1}{2}T=44^{\circ}$. Since $BM\parallel PT$, 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 (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 6 (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 (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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