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

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==Solution 4 (Algebra)==
 
==Solution 4 (Algebra)==
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Let a and b be the distances of R from P and Q, respectively. By the Pythagorean Theorem, we have <math>a^2 + b^2 = 64</math>. Since the area of this triangle is 12, we get <math>a * b = 12 * 2 = 24</math>. Thus <math>b = 24/a</math>. Now substitute this into the other equation to get <math>a^2 + (24/a)^2 = 64</math>. Multiplying by <math>a^2</math> on both sides, we get <math>a^4 + 24 = 64*a^2</math>. Now let <math>y = a^2</math>. Substituting and rearranging, we get <math>y^2 - 64*y + 24 = 0</math>. We reduced this problem to a quadratic equation! By the quadratic formula, our solutions are <math>y = 32 \pm 10*\sqrt{10}</math>. Now substitute back <math>y = a^2</math> to get <math>a = \pm \sqrt{32 \pm 10*\sqrt{10}}</math>. All 4 of these solutions are rational and will work. But our answer is actually <math>4 * 2 = 8</math> as we have only calculated the number of places where R could go above line segment PQ. We need to multiply our 4 by 2 to account for all the places R could go below segment PQ.
+
Let <math>a</math> and <math>b</math> be the distances of <math>R</math> from <math>P</math> and <math>Q,</math> respectively.
 +
 
 +
By the Pythagorean Theorem, we have <cmath>a^2 + b^2 = 64.</cmath>
 +
Since the area of this triangle is <math>12,</math> we get <math>a \cdot b = 12 \cdot 2 = 24.</math> Thus, <math>b = \frac{24}{a}.</math> Now substitute this into the other equation to get <cmath>a^2 + \left(\frac{24}{a}\right)^2 = 64.</cmath>
 +
Multiplying by <math>a^2</math> on both sides, we get <cmath>\text{There is flaw here. I will fix it tomorrow. Bedtime for me. }a^4 + 24 = 64a^2.</cmath>
 +
Now let <math>y = a^2.</math> Substituting and rearranging, we get <cmath>y^2 - 64y + 24 = 0.</cmath>
 +
We reduced this problem to a quadratic equation! By the quadratic formula, our solutions are <math>y = 32 \pm 10\sqrt{10}.</math> Now substitute back <math>y = a^2</math> to get <math>a = \pm \sqrt{32 \pm 10\sqrt{10}}.</math> All <math>4</math> of these solutions are rational and will work. But our answer is actually <math>4\cdot2 = 8</math> as we have only calculated the number of places where R could go above line segment PQ. We need to multiply our 4 by 2 to account for all the places R could go below segment PQ.
 +
 
 
~mewto
 
~mewto
 +
 +
 +
 +
 +
\textbf{(D)}\ 8
  
 
==Video Solution==
 
==Video Solution==

Revision as of 23:37, 29 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 4 (Algebra)

Let $a$ and $b$ be the distances of $R$ from $P$ and $Q,$ respectively.

By the Pythagorean Theorem, we have \[a^2 + b^2 = 64.\] Since the area of this triangle is $12,$ we get $a \cdot b = 12 \cdot 2 = 24.$ Thus, $b = \frac{24}{a}.$ Now substitute this into the other equation to get \[a^2 + \left(\frac{24}{a}\right)^2 = 64.\] Multiplying by $a^2$ on both sides, we get \[\text{There is flaw here. I will fix it tomorrow. Bedtime for me. }a^4 + 24 = 64a^2.\] Now let $y = a^2.$ Substituting and rearranging, we get \[y^2 - 64y + 24 = 0.\] We reduced this problem to a quadratic equation! By the quadratic formula, our solutions are $y = 32 \pm 10\sqrt{10}.$ Now substitute back $y = a^2$ to get $a = \pm \sqrt{32 \pm 10\sqrt{10}}.$ All $4$ of these solutions are rational and will work. But our answer is actually $4\cdot2 = 8$ as we have only calculated the number of places where R could go above line segment PQ. We need to multiply our 4 by 2 to account for all the places R could go below segment PQ.

~mewto



\textbf{(D)}\ 8

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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