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

(→‎Solution 4 (Illustration): Redrew the diagram using Asymptote.)
(Prioritized the solution based on elegance. Let me know if you disagree. Circles are the easiest approach to this problem.)
Line 5: Line 5:
 
<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\  6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12</math>
 
<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\  6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12</math>
  
−
==Solution 1==
+
==Solution 1 (Altitude and Circle)==
−
 
 
−
There are <math>3</math> options here:
 
−
 
 
−
1. <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.
 
−
 
 
−
2. <math>\textbf{Q}</math> is the right angle.
 
−
 
 
−
Using the exact same reasoning, there are also <math>2</math> solutions for this one.
 
−
 
 
−
3. 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 2==
 
−
Note that line segment <math>\overline{PQ}</math> can either be the shorter leg, longer leg or the hypotenuse. If it is the shorter leg, there are two possible points for <math>Q</math> that can satisfy the requirements - that being above or below <math>\overline{PQ}</math>. As such, there are <math>2</math> ways for this case. Similarly, one can find that there are also <math>2</math> ways for point <math>Q</math> to lie if <math>\overline{PQ}</math> is the longer leg. If it is a hypotenuse, then there are <math>4</math> possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is <math>2+2+4=\boxed{\textbf{(D) }8}</math>.
 
−
 
 
−
==Solution 3 (Algebra)==
 
−
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.
 
−
~mewto
 
−
 
 
−
==Solution 4 (Altitude and Circle)==
 
 
Let the brackets denote areas. We know that <cmath>[PQR]=\frac12\cdot PQ\cdot h_R=12.</cmath> Since <math>PQ=8,</math> it follows that <math>h_R=3.</math>
 
Let the brackets denote areas. We know that <cmath>[PQR]=\frac12\cdot PQ\cdot h_R=12.</cmath> Since <math>PQ=8,</math> it follows that <math>h_R=3.</math>
  
Line 129: Line 66:
 
</ol>
 
</ol>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
==Solution 2==
 +
 +
There are <math>3</math> options here:
 +
 +
1. <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.
 +
 +
2. <math>\textbf{Q}</math> is the right angle.
 +
 +
Using the exact same reasoning, there are also <math>2</math> solutions for this one.
 +
 +
3. 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 3==
 +
Note that line segment <math>\overline{PQ}</math> can either be the shorter leg, longer leg or the hypotenuse. If it is the shorter leg, there are two possible points for <math>Q</math> that can satisfy the requirements - that being above or below <math>\overline{PQ}</math>. As such, there are <math>2</math> ways for this case. Similarly, one can find that there are also <math>2</math> ways for point <math>Q</math> to lie if <math>\overline{PQ}</math> is the longer leg. If it is a hypotenuse, then there are <math>4</math> possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is <math>2+2+4=\boxed{\textbf{(D) }8}</math>.
 +
 +
==Solution 4 (Algebra)==
 +
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.
 +
~mewto
  
 
==Video Solution==
 
==Video Solution==

Revision as of 21:02, 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 know 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] For $\triangle PQR,$ note that

  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

There are $3$ options here:

1. $\textbf{P}$ is the right angle.

It's clear that there are $2$ points that fit this, one that's directly to the right of $P$ 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.

2. $\textbf{Q}$ is the right angle.

Using the exact same reasoning, there are also $2$ solutions for this one.

3. 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 $\overline{AB}$ must be $3$ since the area is $12$. From here, we must see if there are valid triangles that satisfy the necessary requirements.

First of all, $\frac{\overline{BC}\cdot\overline{AC}}{2}=12 \implies \overline{BC}\cdot\overline{AC}=24$ because of the area.

Next, $\overline{BC}^2+\overline{AC}^2=64$ from the Pythagorean Theorem.

From here, we must look to see if there are valid solutions. There are multiple ways to do this:

$\textbf{Recognizing min \& max:}$

We know that the minimum value of $\overline{BC}^2+\overline{AC}^2=64$ is when $\overline{BC} = \overline{AC} = \sqrt{24}$. In this case, the equation becomes $24+24=48$, which is LESS than $64$. $\overline{BC}=1, \overline{AC} =24$. The equation becomes $1+576=577$, which is obviously greater than $64$. We can conclude that there are values for $\overline{BC}$ and $\overline{AC}$ in between that satisfy the Pythagorean Theorem.

And since $\overline{BC} \neq \overline{AC}$, the triangle is not isoceles, meaning we could reflect it over $\overline{AB}$ and/or the line perpendicular to $\overline{AB}$ for a total of $4$ triangles this case.

Therefore, the answer is $2+2+4=\boxed{\textbf{(D) }8}$.

Solution 3

Note that line segment $\overline{PQ}$ can either be the shorter leg, longer leg or the hypotenuse. If it is the shorter leg, there are two possible points for $Q$ that can satisfy the requirements - that being above or below $\overline{PQ}$. As such, there are $2$ ways for this case. Similarly, one can find that there are also $2$ ways for point $Q$ to lie if $\overline{PQ}$ is the longer leg. If it is a hypotenuse, then there are $4$ possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is $2+2+4=\boxed{\textbf{(D) }8}$.

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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