Difference between revisions of "2024 IMO Problems/Problem 4"
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Prove that <math>\angle KIL + \angle YPX = 180^{\circ}</math> | Prove that <math>\angle KIL + \angle YPX = 180^{\circ}</math> | ||
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=QphkkutmY5M&t=1s | ||
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/WnZv3fdpFXo | https://youtu.be/WnZv3fdpFXo |
Revision as of 23:58, 26 July 2024
Let be a triangle with
. Let the incentre and incircle of triangle
be
and
, respectively. Let
be the point on line
different from
such that the line
through
parallel to
is tangent to
. Similarly, let
be the point on line
different from
such that the line through
parallel to
is tangent to
. Let
intersect the circumcircle of
triangle
again at
. Let
and
be the midpoints of
and
, respectively.
Prove that
.
Video Solution
https://www.youtube.com/watch?v=QphkkutmY5M&t=1s
Video Solution
Video Solution
Part 1: Derive tangent values and
with trig values of angles
,
,
Part 2: Derive tangent values and
with side lengths
,
,
, where
is the midpoint of
Part 3: Prove that and
.
Comments: Although this is an IMO problem, the skills needed to solve this problem have all previously tested in AMC and its system math contests, such as HMMT.
Evidence 1: 2020 Spring HMMT Geometry Round Problem 8
I used the property that because point is on the angle bisector
,
is isosceles. This is a crucial step to analyze
. This technique was previously tested in this HMMT problem.
Evidence 2: 2022 AMC 12A Problem 25
The technique in this AMC problem can be easily and directly applied to this IMO problem to quickly determine the locations of points and
. If you read my solutions to both this AMC problem and this IMO problem, you will find that I simply took exactly the same approach to solve both.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)