Difference between revisions of "2022 AMC 10A Problems/Problem 13"
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MRENTHUSIASM (talk | contribs) (→Diagram) |
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<math>\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12</math> | <math>\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12</math> | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(300); | ||
+ | real r = 4*sqrt(114)/13; | ||
+ | pair A, B, C, D, P, X, Y; | ||
+ | A = origin; | ||
+ | B = (2,r); | ||
+ | C = (3/2*sqrt(2^2+r^2),0); | ||
+ | D = A + 2*(C-B); | ||
+ | P = B + 2*dir(C-B); | ||
+ | X = intersectionpoint(B--D,A--P); | ||
+ | Y = intersectionpoint(B--D,A--C); | ||
+ | dot("$A$",A,1.5*W,linewidth(4)); | ||
+ | dot("$B$",B,1.5*N,linewidth(4)); | ||
+ | dot("$C$",C,1.5*E,linewidth(4)); | ||
+ | dot("$P$",P,1.5*dir(P),linewidth(4)); | ||
+ | dot("$D$",D,1.5*dir(D),linewidth(4)); | ||
+ | dot(X^^Y,linewidth(4)); | ||
+ | markscalefactor=0.03; | ||
+ | draw(rightanglemark(B,X,A),red); | ||
+ | draw(anglemark(P,A,B,20), red); | ||
+ | draw(anglemark(C,A,P,20), red); | ||
+ | add(pathticks(anglemark(P,A,B,20), n = 1, r = 0.1, s = 7, red)); | ||
+ | add(pathticks(anglemark(C,A,P,20), n = 1, r = 0.1, s = 7, red)); | ||
+ | draw(A--B--C--cycle^^A--P^^B--D^^A--D); | ||
+ | draw(B--C,MidArrow(0.3cm,Fill(red))); | ||
+ | draw(A--D,MidArrow(0.3cm,Fill(red))); | ||
+ | label("$2$",midpoint(B--P),rotate(90)*dir(midpoint(P--B)--P),red); | ||
+ | label("$3$",midpoint(P--C),rotate(90)*dir(midpoint(C--P)--C),red); | ||
+ | </asy> | ||
+ | ~MRENTHUSIASM | ||
==Solution 1 (Angle Bisector Theorem and Similar Triangles)== | ==Solution 1 (Angle Bisector Theorem and Similar Triangles)== |
Latest revision as of 04:47, 5 November 2024
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Angle Bisector Theorem and Similar Triangles)
- 4 Solution 2 (Auxiliary Lines)
- 5 Solution 3 (Slopes)
- 6 Solution 4 (Assumption)
- 7 Video Solution 1
- 8 Video Solution 2
- 9 Video Solution 3
- 10 Video Solution 4
- 11 Video Solution 5 by SpreadTheMathLove
- 12 Video Solution 6 by Lucas637
- 13 See Also
Problem
Let be a scalene triangle. Point lies on so that bisects The line through perpendicular to intersects the line through parallel to at point Suppose and What is
Diagram
~MRENTHUSIASM
Solution 1 (Angle Bisector Theorem and Similar Triangles)
Suppose that intersects and at and respectively. By Angle-Side-Angle, we conclude that
Let By the Angle Bisector Theorem, we have or
By alternate interior angles, we get and Note that by the Angle-Angle Similarity, with the ratio of similitude It follows that
~MRENTHUSIASM
Solution 2 (Auxiliary Lines)
Let the intersection of and be , and the intersection of and be . Draw a line from to , and label the point of intersection .
By adding this extra line, we now have many pairs of similar triangles. We have , with a ratio of , so and . We also have with ratio . Additionally, (with an unknown ratio). It is also true that .
Suppose the area of is . Then, . Because and share the same height and have a base ratio of , . Because and share the same height and have a base ratio of , , , and thus . Thus, .
Finally, we have , and because these triangles share the same height . Notice that these side lengths are corresponding side lengths of the similar triangles and . This means that .
~mathboy100
Solution 3 (Slopes)
Let point be the origin, with being on the positive -axis and being in the first quadrant.
By the Angle Bisector Theorem, . Thus, assume that , and .
Let the perpendicular from to be .
Using Heron's formula,
Hence,
Next, we have
The slope of line is thus
Therefore, since the slopes of perpendicular lines have a product of , the slope of line is . This means that we can solve for the coordinates of :
We also know that the coordinates of are , because and .
Since the -coordinates of and are the same, and their -coordinates differ by , the distance between them is . Our answer is
~mathboy100
Solution 4 (Assumption)
Since there is only one possible value of , we assume . By the angle bisector theorem, , so and . Now observe that . Let the intersection of and be . Then . Consequently, and therefore , so , and we're done!
Video Solution 1
~Education, the Study of Everything
Video Solution 2
- Whiz
Video Solution 3
Video Solution 4
Video Solution 5 by SpreadTheMathLove
https://www.youtube.com/watch?v=nhlpSATltRU
~Ismail.maths93
Video Solution 6 by Lucas637
https://www.youtube.com/watch?v=R1CtcZ2pWVk
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.