Difference between revisions of "2020 AMC 10B Problems/Problem 8"
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<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 (Geometry)== |
+ | 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> | ||
− | + | We construct a circle with diameter <math>\overline{PQ}.</math> All such locations for <math>R</math> are shown below: | |
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<asy> | <asy> | ||
− | pair | + | /* 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); | |
− | draw( | + | R8 = (4,-3); |
− | + | path C; | |
− | + | C = Circle(O,4); | |
− | + | R3 = intersectionpoints(C,R1--R4)[0]; | |
− | + | R2 = intersectionpoints(C,R1--R4)[1]; | |
− | dot("$ | + | R6 = intersectionpoints(C,R5--R8)[0]; |
− | dot("$ | + | R7 = intersectionpoints(C,R5--R8)[1]; |
− | dot("$ | + | 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); | |
− | Label | + | 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> | </asy> | ||
− | + | We apply casework to the right angle of <math>\triangle PQR:</math> | |
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>If <math>\angle P=90^\circ,</math> then <math>R\in\{R_1,R_5\}</math> by the tangent.</li><p> | ||
+ | <li>If <math>\angle Q=90^\circ,</math> then <math>R\in\{R_4,R_8\}</math> by the tangent.</li><p> | ||
+ | <li>If <math>\angle R=90^\circ,</math> then <math>R\in\{R_2,R_3,R_6,R_7\}</math> by the Inscribed Angle Theorem.</li><p> | ||
+ | </ol> | ||
+ | Together, there are <math>\boxed{\textbf{(D)}\ 8}</math> such locations for <math>R.</math> | ||
− | + | <u><b>Remarks</b></u> | |
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>The reflections of <math>R_1,R_2,R_3,R_4</math> about <math>\overleftrightarrow{PQ}</math> are <math>R_5,R_6,R_7,R_8,</math> respectively.</li><p> | ||
+ | <li>The reflections of <math>R_1,R_2,R_5,R_6</math> about the perpendicular bisector of <math>\overline{PQ}</math> are <math>R_4,R_3,R_8,R_7,</math> respectively.</li><p> | ||
+ | </ol> | ||
+ | ~MRENTHUSIASM | ||
− | + | ==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> | |
− | <math>\textbf{ | + | 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 | |
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− | + | ==Video Solution (HOW TO CRITICALLY THINK!!!)== | |
+ | https://youtu.be/C_9Wa_owu9s | ||
− | + | ~Education, the Study of Everything | |
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/OHR_6U686Qg | https://youtu.be/OHR_6U686Qg | ||
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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
Contents
Problem
Points and lie in a plane with . How many locations for point in this plane are there such that the triangle with vertices , , and is a right triangle with area square units?
Solution 1 (Geometry)
Let the brackets denote areas. We are given that Since it follows that
We construct a circle with diameter All such locations for are shown below:
We apply casework to the right angle of
- If then by the tangent.
- If then by the tangent.
- If then by the Inscribed Angle Theorem.
Together, there are such locations for
Remarks
- The reflections of about are respectively.
- The reflections of about the perpendicular bisector of are respectively.
~MRENTHUSIASM
Solution 2 (Algebra)
Let the brackets denote areas. We are given that Since it follows that
Without the loss of generality, let and We conclude that the -coordinate of must be
We apply casework to the right angle of
-
The -coordinate of must be so we have
In this case, there are such locations for
-
The -coordinate of must be so we have
In this case, there are such locations for
-
For the Pythagorean Theorem gives Solving this equation, we have or
For we have by a similar process.
In this case, there are such locations for
Together, there are such locations for
~MRENTHUSIASM ~mewto
Video Solution (HOW TO CRITICALLY THINK!!!)
~Education, the Study of Everything
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.