Difference between revisions of "2022 AMC 10A Problems/Problem 13"
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Let <math>AB=AY=2x.</math> By the Angle Bisector Theorem, we have <math>AC=3x,</math> or <math>YC=x.</math> | Let <math>AB=AY=2x.</math> By the Angle Bisector Theorem, we have <math>AC=3x,</math> or <math>YC=x.</math> | ||
− | By parallel lines, we get <math>\angle YAD=\angle YCB</math> and <math>\angle YDA=\angle YBC.</math> Note that <math>\triangle ADY \sim \triangle CBY</math> by the Angle-Angle Similarity, with the ratio of similitude <math>\ | + | By parallel lines, we get <math>\angle YAD=\angle YCB</math> and <math>\angle YDA=\angle YBC.</math> Note that <math>\triangle ADY \sim \triangle CBY</math> by the Angle-Angle Similarity, with the ratio of similitude <math>\frac{AY}{CY}=2.</math> It follows that <math>AD=2CB=2(BP+PC)=\boxed{\textbf{(C) } 10}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 23:16, 11 November 2022
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
Solution
DIAGRAM IN PROGRESS.
WILL BE DONE TOMORROW, WAIT FOR ME THANKS.
Suppose that intersect
and
at
and
respectively. By Angle-Side-Angle, we conclude that
Let By the Angle Bisector Theorem, we have
or
By parallel lines, we get and
Note that
by the Angle-Angle Similarity, with the ratio of similitude
It follows that
~MRENTHUSIASM
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.