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Difference between revisions of "2020 AMC 10B Problems/Problem 8"

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(→‎Video Solution)
 
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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>
  
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There are <math>3</math> options here:
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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>
  
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2. <math>\textbf{Q}</math> is the right angle.  
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<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 the 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 temprarily 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
  
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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
  
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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>
 
  
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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>.
+
https://youtu.be/cUzK5DqKaRY
−
<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.
 
  
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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.
+
~savannahsolver
−
 
 
−
==Solution 2==
 
−
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==
 
−
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 14:00, 12 July 2024

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 the 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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