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

m (→‎Solution 1 (Geometry))
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~MRENTHUSIASM ~mewto
 
~MRENTHUSIASM ~mewto
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==Solution 3 (Imagination and logic)==
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We can draw out line <math>PQ</math> and see that <math>PQ</math> must either be the base or the height of the right triangle. We can now split this problem in 2 cases.
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 +
Case 1: <math>PQ</math> is the height. We can have point 2 on each side of <math>Q</math> and each side of <math>P</math>. Which leads to <math>4</math> cases.
 +
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Case 2: <math>PQ</math> is the base. We can see that R can be a point below <math>P</math> or above and the same for <math>Q</math> Which again leads to <math>4</math> cases.
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<math>4+4=\boxed{\textbf{(D)}\ 8}</math> locations for <math>R.</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 20:09, 4 November 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(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

Solution 3 (Imagination and logic)

We can draw out line $PQ$ and see that $PQ$ must either be the base or the height of the right triangle. We can now split this problem in 2 cases.

Case 1: $PQ$ is the height. We can have point 2 on each side of $Q$ and each side of $P$. Which leads to $4$ cases.

Case 2: $PQ$ is the base. We can see that R can be a point below $P$ or above and the same for $Q$ Which again leads to $4$ cases.

$4+4=\boxed{\textbf{(D)}\ 8}$ locations for $R.$

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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