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

(→‎Solution (Olympiad nuke))
(Solutions should be first come first serve. So, I moved Solution 6 to the end.)
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== Problem ==
 
== Problem ==
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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>
 
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>
  
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<math>\textbf{(A)} 76 \qquad  
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<math>\textbf{(A) } 76 \qquad  
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\textbf{(B)} 77 \qquad  
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\textbf{(B) } 77 \qquad  
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\textbf{(C)} 78 \qquad  
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\textbf{(C) } 78 \qquad  
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\textbf{(D)} 79 \qquad  
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\textbf{(D) } 79 \qquad  
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\textbf{(E)} 80 </math>
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\textbf{(E) } 80 </math>
  
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== Solution (Olympiad nuke) ==
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== Solution 1 ==
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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{E}.</math>
 
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== Solution ==
 
  
 
Let <math>P</math> be the origin, and <math>PA</math> lie on the x axis.
 
Let <math>P</math> be the origin, and <math>PA</math> lie on the x axis.
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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)
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so the answer is <math>\boxed{\textbf{(E) } 80}.</math>
  
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Note: Though this solution is excellent, setting <math>M = (0,0)</math> makes life a tad bit easier ~ MathleteMA
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~lifeisgood03
 +
 
 +
Note: Though this solution is excellent, setting <math>M = (0,0)</math> makes life a tad bit easier
 +
 
 +
~MathleteMA
  
 
==Solution 2 (Overkill)==
 
==Solution 2 (Overkill)==
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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>.  
 
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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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>.
+
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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(stronto)
+
~stronto
 +
 
 +
<u><b>Sidenote</b></u>
  
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===Sidenote===
 
 
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.
  
 
==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) } 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>
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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>).
  
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(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!
  
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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>.  
 
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>.  
 
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>.  
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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>.
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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>.
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(Surefire2019)
+
 
 +
~Surefire2019
  
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==Solution 5 (Simplest, I think)==
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==Solution 5 (Simplest, I think?)==
  
 
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>.  
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Therefore, <math>\angle QAP=\angle QAG+\angle TAP=56^{\circ}+44^{\circ}=100^{\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{80^{\circ}}</math>
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Since we are asked for the acute angle between the two lines, the answer to this problem is <math>\boxed{\textbf{(E) } 80}</math>
  
 
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)
 
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)
  
−
=== Video Solution by Richard Rusczyk ===
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== Solution 6 (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 ==
  
 
https://artofproblemsolving.com/videos/amc/2018amc12a/473
 
https://artofproblemsolving.com/videos/amc/2018amc12a/473

Revision as of 12:16, 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 $PA?$

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

Solution 1

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)

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

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 $ND//BC$ 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 $BM//PT$, 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 (Simplest, I think?)

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

$\because M$, $N$ is the middle point of $AP$ and $QU$

$\therefore MN$ is the middle line 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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