2022 AMC 10A Problems/Problem 13
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.