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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.)
(→‎Video Solution)
 
(11 intermediate revisions by 5 users not shown)
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We construct a circle with diameter <math>\overline{PQ}.</math> All such locations for <math>R</math> are shown below:
 
We construct a circle with diameter <math>\overline{PQ}.</math> All such locations for <math>R</math> are shown below:
 +
 
<asy>
 
<asy>
 
/* Made by MRENTHUSIASM */
 
/* Made by MRENTHUSIASM */
Line 45: Line 46:
 
dot(I1,linewidth(4));
 
dot(I1,linewidth(4));
 
dot(I2,linewidth(4));
 
dot(I2,linewidth(4));
−
 
+
Label L1 = Label("$8$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
−
Label L1 = Label("$8$", align=(0,0), position=MidPoint, filltype=Fill(white));
+
Label L2 = Label("$3$", align=(0,0), position=MidPoint, filltype=Fill(0,3,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(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(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));
 
draw(Q+(2,0)--R8+(2,0), L=L2, arrow=Arrows(),bar=Bars(15));
 
</asy>
 
</asy>
 +
 
We apply casework to the right angle of <math>\triangle PQR:</math>
 
We apply casework to the right angle of <math>\triangle PQR:</math>
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
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==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>
 +
<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>
  
 
~MRENTHUSIASM ~mewto
 
~MRENTHUSIASM ~mewto
 +
 +
==Video Solution (HOW TO CRITICALLY THINK!!!)==
 +
https://youtu.be/C_9Wa_owu9s
 +
 +
~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/OHR_6U686Qg
 
https://youtu.be/OHR_6U686Qg
−
 
−
~IceMatrix
 
  
 
https://youtu.be/cUzK5DqKaRY
 
https://youtu.be/cUzK5DqKaRY
  
 
~savannahsolver
 
~savannahsolver
 +
 +
== 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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