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

Revision as of 22:55, 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 (Altitude and Circle)

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 $\triangle PQR:$

If $\angle P=90^\circ,$ then $R\in\{R_1,R_5\}$ by the tangent.

If $\angle Q=90^\circ,$ then $R\in\{R_4,R_8\}$ by the tangent.

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

The reflections of $R_1,R_2,R_3,R_4$ about $\overleftrightarrow{PQ}$ are $R_5,R_6,R_7,R_8,$ respectively.

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 $a * b = 12 * 2 = 24$ . Thus $b = 24/a$ . Now substitute this into the other equation to get $a^2 + (24/a)^2 = 64$ . Multiplying by $a^2$ on both sides, we get $a^4 + 24 = 64*a^2$ . Now let $y = a^2$ . Substituting and rearranging, we get $y^2 - 64*y + 24 = 0$ . 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 $4 * 2 = 8$ 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

Video Solution

https://youtu.be/OHR_6U686Qg

~IceMatrix

https://youtu.be/cUzK5DqKaRY

~savannahsolver

@@ Line 66: / Line 66: @@
 </ol>
 ~MRENTHUSIASM
-==Solution 2==
-There are <math>3</math> options here:
-. <math>\textbf{P}</math> is the right angle.
-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.
-. <math>\textbf{Q}</math> is the right angle.
-Using the exact same reasoning, there are also <math>2</math> solutions for this one.
-. The new point is the right angle.
-<asy>
-pair  A, B, C, D, X, Y;
-A = (0,0);
-B = (0,8);
-C = (3,6.64575131106);
-D = (0,6.64575131106);
-X = (0,4);
-Y = (1.5,6.64575131106);
-draw(A--B--C--A);
-draw(C--D);
-label("$8$", X, W);
-label("$3$", Y, S);
-dot("$A$", A, S);
-dot("$B$", B, N);
-dot("$C$", C, E);
-draw(rightanglemark(A, C, B));
-draw(rightanglemark(A, D, C));
-Label AB= Label("$8$", position=MidPoint);
-</asy>
-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.
-First of all, <math>\frac{\overline{BC}\cdot\overline{AC}}{2}=12 \implies \overline{BC}\cdot\overline{AC}=24</math> because of the area.
-Next, <math>\overline{BC}^2+\overline{AC}^2=64</math> from the Pythagorean Theorem.
-From here, we must look to see if there are valid solutions. There are multiple ways to do this:
-<math>\textbf{Recognizing min \& max:}</math>
-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>.
-<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.
-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.
-Therefore, the answer is <math>2+2+4=\boxed{\textbf{(D) }8}</math>.
 ==Solution 4 (Algebra)==

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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