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

(→‎Solution 4 (Algebra): I found some flaws in this solution too. First, the equation was not solved correctly. Secondly, we should do casework which one is the right angle before we apply the Pythagorean Theorem.)
(Polished the algebra solution.)
Line 68: Line 68:
  
 
==Solution 2 (Algebra)==
 
==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>
 +
 +
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>
 +
 +
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>
 +
In this case, there are <math>2</math> such locations for <math>R.</math><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>
 +
In this case, there are <math>2</math> such locations for <math>R.</math><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>
 +
In this case, there are <math>4</math> such locations for <math>R.</math><p></li>
 +
</ol>
 +
Together, there are <math>2+2+4=\boxed{\textbf{(D)}\ 8}</math> such locations for <math>R.</math>
  
 
~MRENTHUSIASM ~mewto
 
~MRENTHUSIASM ~mewto

Revision as of 07:44, 30 September 2021

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(white)); Label L2 = Label("$3$", align=(0,0), position=MidPoint, filltype=Fill(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 $2$ such locations for $R.$

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

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

    In this case, there are $2$ such locations for $R.$

  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 $4$ such locations for $R.$

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

~MRENTHUSIASM ~mewto

Video Solution

https://youtu.be/OHR_6U686Qg

~IceMatrix

https://youtu.be/cUzK5DqKaRY

~savannahsolver

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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