Difference between revisions of "2022 AMC 10A Problems/Problem 13"
(→Solution 3 (Cheaty solution if you are almost out of time)) |
(→Solution 3 (Cheaty solution if you are almost out of time)) |
||
Line 19: | Line 19: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Solution 3 ( | + | == Solution 3 (Cheese Solution) == |
<asy> | <asy> | ||
size(300); | size(300); |
Revision as of 02:32, 15 November 2022
Contents
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
Solution 3 (Cheese Solution)
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 by OmegaLearn
~ pi_is_3.14
Video Solution (Quick and Simple)
~Education, the Study of Everything
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.