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2020 AMC 10B Problems/Problem 8: Difference between revisions

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<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\  6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12</math>
<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\  6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12</math>


==Solution 1==
==Solution 1 (Geometry)==
Let the brackets denote areas. We are given that <cmath>[PQR]=\frac12\cdot PQ\cdot h_R=12.</cmath> Since <math>PQ=8,</math> it follows that <math>h_R=3.</math>


There are <math>3</math> options here:
We construct a circle with diameter <math>\overline{PQ}.</math> All such locations for <math>R</math> are shown below:


1. <math>\textbf{P}</math> is the right angle.
<asy>
/* Made by MRENTHUSIASM */
size(250);
pair O, P, Q, R1, R2, R3, R4, R5, R6, R7, R8, I1, I2;
O = (0,0);
P = (-4,0);
Q = (4,0);
R1 = (-4,3);
R4 = (4,3);
R5 = (-4,-3);
R8 = (4,-3);
path C;
C = Circle(O,4);
R3 = intersectionpoints(C,R1--R4)[0];
R2 = intersectionpoints(C,R1--R4)[1];
R6 = intersectionpoints(C,R5--R8)[0];
R7 = intersectionpoints(C,R5--R8)[1];
I1 = intersectionpoint(R2--R6,P--Q);
I2 = intersectionpoint(R3--R7,P--Q);
markscalefactor=0.0375;
draw(rightanglemark(R1,P,Q)^^rightanglemark(R2,I1,Q)^^rightanglemark(R3,I2,P)^^rightanglemark(R4,Q,P),red);
draw(Circle(O,4),dashed);
draw(R1--R5^^R4--R8^^R2--R6^^R3--R7^^P--Q);
dot(O,linewidth(4));
dot("$P$",P,1.5W,linewidth(4));
dot("$Q$",Q,1.5E,linewidth(4));
dot("$R_1$",R1,1.5NW,blue+linewidth(4));
dot("$R_4$",R4,1.5NE,blue+linewidth(4));
dot("$R_5$",R5,1.5SW,blue+linewidth(4));
dot("$R_8$",R8,1.5SE,blue+linewidth(4));
dot("$R_2$",R2,1.5NW,blue+linewidth(4));
dot("$R_3$",R3,1.5NE,blue+linewidth(4));
dot("$R_6$",R6,1.5SW,blue+linewidth(4));
dot("$R_7$",R7,1.5SE,blue+linewidth(4));
dot(I1,linewidth(4));
dot(I2,linewidth(4));
Label L1 = Label("$8$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
Label L2 = Label("$3$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white));
draw(P-(0,5)--Q-(0,5), L=L1, arrow=Arrows(),bar=Bars(15));
draw(R4+(2,0)--Q+(2,0), L=L2, arrow=Arrows(),bar=Bars(15));
draw(Q+(2,0)--R8+(2,0), L=L2, arrow=Arrows(),bar=Bars(15));
</asy>


It's clear that there are <math>2</math> points that fit this, one that's directly to the right of <math>P</math> and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is.
We apply casework to the right angle of <math>\triangle PQR:</math>
<ol style="margin-left: 1.5em;">
  <li>If <math>\angle P=90^\circ,</math> then <math>R\in\{R_1,R_5\}</math> by the tangent.</li><p>
  <li>If <math>\angle Q=90^\circ,</math> then <math>R\in\{R_4,R_8\}</math> by the tangent.</li><p>
  <li>If <math>\angle R=90^\circ,</math> then <math>R\in\{R_2,R_3,R_6,R_7\}</math> by the Inscribed Angle Theorem.</li><p>
</ol>
Together, there are <math>\boxed{\textbf{(D)}\ 8}</math> such locations for <math>R.</math>


2. <math>\textbf{Q}</math> is the right angle.  
<u><b>Remarks</b></u>
<ol style="margin-left: 1.5em;">
  <li>The reflections of <math>R_1,R_2,R_3,R_4</math> about <math>\overleftrightarrow{PQ}</math> are <math>R_5,R_6,R_7,R_8,</math> respectively.</li><p>
  <li>The reflections of <math>R_1,R_2,R_5,R_6</math> about the perpendicular bisector of <math>\overline{PQ}</math> are <math>R_4,R_3,R_8,R_7,</math> respectively.</li><p>
</ol>
~MRENTHUSIASM


Using the exact same reasoning, there are also <math>2</math> solutions for this one.
==Solution 2 (Algebra)==
Let the brackets denote areas. We are given that <cmath>[PQR]=\frac12\cdot PQ\cdot h_R=12.</cmath> Since <math>PQ=8,</math> it follows that <math>h_R=3.</math>


3. The new point is the right angle.  
Without loss of generality, let <math>P=(-4,0)</math> and <math>Q=(4,0).</math> We conclude that the <math>y</math>-coordinate of <math>R</math> must be <math>\pm3.</math>


(Diagram temporarily removed due to asymptote error)
We apply casework to the right angle of <math>\triangle PQR:</math>
<ol style="margin-left: 1.5em;">
  <li><math>\angle P=90^\circ.</math> <p>
The <math>x</math>-coordinate of <math>R</math> must be <math>-4,</math> so we have <math>R=(-4,\pm3).</math> <p>
<b>In this case, there are <math>\boldsymbol{2}</math> such locations for <math>\boldsymbol{R.}</math></b> <p></li>
  <li><math>\angle Q=90^\circ.</math> <p>
The <math>x</math>-coordinate of <math>R</math> must be <math>4,</math> so we have <math>R=(4,\pm3).</math> <p>
<b>In this case, there are <math>\boldsymbol{2}</math> such locations for <math>\boldsymbol{R.}</math></b> <p></li>
  <li><math>\angle R=90^\circ.</math> <p>
For <math>R=(x,3),</math> the Pythagorean Theorem <math>PR^2+QR^2=PQ^2</math> gives <cmath>\left[(x+4)^2+3^2\right]+\left[(x-4)^2+3^2\right]=8^2.</cmath> Solving this equation, we have <math>x=\pm\sqrt7,</math> or <math>R=\left(\pm\sqrt7,3\right).</math> <p>
For <math>R=(x,-3),</math> we have <math>R=\left(\pm\sqrt7,-3\right)</math> by a similar process. <p>
<b>In this case, there are <math>\boldsymbol{4}</math> such locations for <math>\boldsymbol{R.}</math></b> <p></li>
</ol>
Together, there are <math>2+2+4=\boxed{\textbf{(D)}\ 8}</math> such locations for <math>R.</math>


The diagram looks something like this. We know that the altitude to base <math>\overline{AB}</math> must be <math>3</math> since the area is <math>12</math>. From here, we must see if there are valid triangles that satisfy the necessary requirements.
~MRENTHUSIASM ~mewto


First of all, <math>\frac{\overline{BC}\cdot\overline{AC}}{2}=12 \implies \overline{BC}\cdot\overline{AC}=24</math> because of the area.
==Video Solution (HOW TO CRITICALLY THINK!!!)==
https://youtu.be/C_9Wa_owu9s


Next, <math>\overline{BC}^2+\overline{AC}^2=64</math> from the Pythagorean Theorem.
~Education, the Study of Everything


From here, we must look to see if there are valid solutions. There are multiple ways to do this:
==Video Solution==
 
https://youtu.be/OHR_6U686Qg
<math>\textbf{Recognizing min \& max:}</math>
 
We know that the minimum value of <math>\overline{BC}^2+\overline{AC}^2=64</math> is when <math>\overline{BC} = \overline{AC} = \sqrt{24}</math>. In this case, the equation becomes <math>24+24=48</math>, which is LESS than <math>64</math>.
<math>\overline{BC}=1, \overline{AC} =24</math>. The equation becomes <math>1+576=577</math>, which is obviously greater than <math>64</math>. We can conclude that there are values for <math>\overline{BC}</math> and <math>\overline{AC}</math> in between that satisfy the Pythagorean Theorem.
 
And since  <math>\overline{BC} \neq \overline{AC}</math>, the triangle is not isoceles, meaning we could reflect it over <math>\overline{AB}</math> and/or the line perpendicular to <math>\overline{AB}</math> for a total of <math>4</math> triangles this case.
 
Therefore, the answer is <math>2+2+4=\boxed{\textbf{(D) }8}</math>.


==Solution 2==
https://youtu.be/cUzK5DqKaRY
Note that line segment <math>\overline{PQ}</math> can either be the shorter leg, longer leg or the hypotenuse. If it is the shorter leg, there are two possible points for <math>Q</math> that can satisfy the requirements - that being above or below <math>\overline{PQ}</math>. As such, there are <math>2</math> ways for this case. Similarly, one can find that there are also <math>2</math> ways for point <math>Q</math> to lie if <math>\overline{PQ}</math> is the longer leg. If it is a hypotenuse, then there are <math>4</math> possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is <math>2+2+4=\boxed{\textbf{(D) }8}</math>.


==Video Solution==
~savannahsolver
https://youtu.be/OHR_6U686Qg


~IceMatrix
== Video Solution by Sohil Rathi==
https://youtu.be/GrCtzL0S-Uo?t=19


== See Also ==
== See Also ==

Latest revision as of 10:54, 22 October 2025

Problem

Points $P$ and $Q$ lie in a plane with $PQ=8$. How many locations for point $R$ in this plane are there such that the triangle with vertices $P$, $Q$, and $R$ is a right triangle with area $12$ square units?

$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$

Solution 1 (Geometry)

Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$

We construct a circle with diameter $\overline{PQ}.$ All such locations for $R$ are shown below:

[asy] /* Made by MRENTHUSIASM */ size(250); pair O, P, Q, R1, R2, R3, R4, R5, R6, R7, R8, I1, I2; O = (0,0); P = (-4,0); Q = (4,0); R1 = (-4,3); R4 = (4,3); R5 = (-4,-3); R8 = (4,-3); path C; C = Circle(O,4); R3 = intersectionpoints(C,R1--R4)[0]; R2 = intersectionpoints(C,R1--R4)[1]; R6 = intersectionpoints(C,R5--R8)[0]; R7 = intersectionpoints(C,R5--R8)[1]; I1 = intersectionpoint(R2--R6,P--Q); I2 = intersectionpoint(R3--R7,P--Q); markscalefactor=0.0375; draw(rightanglemark(R1,P,Q)^^rightanglemark(R2,I1,Q)^^rightanglemark(R3,I2,P)^^rightanglemark(R4,Q,P),red); draw(Circle(O,4),dashed); draw(R1--R5^^R4--R8^^R2--R6^^R3--R7^^P--Q); dot(O,linewidth(4)); dot("$P$",P,1.5W,linewidth(4)); dot("$Q$",Q,1.5E,linewidth(4)); dot("$R_1$",R1,1.5NW,blue+linewidth(4)); dot("$R_4$",R4,1.5NE,blue+linewidth(4)); dot("$R_5$",R5,1.5SW,blue+linewidth(4)); dot("$R_8$",R8,1.5SE,blue+linewidth(4)); dot("$R_2$",R2,1.5NW,blue+linewidth(4)); dot("$R_3$",R3,1.5NE,blue+linewidth(4)); dot("$R_6$",R6,1.5SW,blue+linewidth(4)); dot("$R_7$",R7,1.5SE,blue+linewidth(4)); dot(I1,linewidth(4)); dot(I2,linewidth(4)); Label L1 = Label("$8$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$3$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(P-(0,5)--Q-(0,5), L=L1, arrow=Arrows(),bar=Bars(15)); draw(R4+(2,0)--Q+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); draw(Q+(2,0)--R8+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy]

We apply casework to the right angle of $\triangle PQR:$

  1. If $\angle P=90^\circ,$ then $R\in\{R_1,R_5\}$ by the tangent.

  2. If $\angle Q=90^\circ,$ then $R\in\{R_4,R_8\}$ by the tangent.

  3. If $\angle R=90^\circ,$ then $R\in\{R_2,R_3,R_6,R_7\}$ by the Inscribed Angle Theorem.

Together, there are $\boxed{\textbf{(D)}\ 8}$ such locations for $R.$

Remarks

  1. The reflections of $R_1,R_2,R_3,R_4$ about $\overleftrightarrow{PQ}$ are $R_5,R_6,R_7,R_8,$ respectively.

  2. The reflections of $R_1,R_2,R_5,R_6$ about the perpendicular bisector of $\overline{PQ}$ are $R_4,R_3,R_8,R_7,$ respectively.

~MRENTHUSIASM

Solution 2 (Algebra)

Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$

Without loss of generality, let $P=(-4,0)$ and $Q=(4,0).$ We conclude that the $y$-coordinate of $R$ must be $\pm3.$

We apply casework to the right angle of $\triangle PQR:$

  1. $\angle P=90^\circ.$

    The $x$-coordinate of $R$ must be $-4,$ so we have $R=(-4,\pm3).$

    In this case, there are $\boldsymbol{2}$ such locations for $\boldsymbol{R.}$

  2. $\angle Q=90^\circ.$

    The $x$-coordinate of $R$ must be $4,$ so we have $R=(4,\pm3).$

    In this case, there are $\boldsymbol{2}$ such locations for $\boldsymbol{R.}$

  3. $\angle R=90^\circ.$

    For $R=(x,3),$ the Pythagorean Theorem $PR^2+QR^2=PQ^2$ gives \[\left[(x+4)^2+3^2\right]+\left[(x-4)^2+3^2\right]=8^2.\] Solving this equation, we have $x=\pm\sqrt7,$ or $R=\left(\pm\sqrt7,3\right).$

    For $R=(x,-3),$ we have $R=\left(\pm\sqrt7,-3\right)$ by a similar process.

    In this case, there are $\boldsymbol{4}$ such locations for $\boldsymbol{R.}$

Together, there are $2+2+4=\boxed{\textbf{(D)}\ 8}$ such locations for $R.$

~MRENTHUSIASM ~mewto

Video Solution (HOW TO CRITICALLY THINK!!!)

https://youtu.be/C_9Wa_owu9s

~Education, the Study of Everything

Video Solution

https://youtu.be/OHR_6U686Qg

https://youtu.be/cUzK5DqKaRY

~savannahsolver

Video Solution by Sohil Rathi

https://youtu.be/GrCtzL0S-Uo?t=19

See Also

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 10 Problems and Solutions

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