Difference between revisions of "2024 IMO Problems/Problem 4"
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==Video Solution== | ==Video Solution== | ||
Part 1: Derive tangent values <math>\angle AIL</math> and <math>\angle AIK</math> with trig values of angles <math>\frac{A}{2}</math>, <math>\frac{B}{2}</math>, <math>\frac{C}{2}</math> | Part 1: Derive tangent values <math>\angle AIL</math> and <math>\angle AIK</math> with trig values of angles <math>\frac{A}{2}</math>, <math>\frac{B}{2}</math>, <math>\frac{C}{2}</math> | ||
+ | |||
https://youtu.be/p_AmooMMln4 | https://youtu.be/p_AmooMMln4 | ||
Part 2: Derive tangent values <math>\angle XPM</math> and <math>\angle YPM</math> with side lengths <math>AB</math>, <math>BC</math>, <math>CA</math>, where <math>M</math> is the midpoint of <math>BC</math> | Part 2: Derive tangent values <math>\angle XPM</math> and <math>\angle YPM</math> with side lengths <math>AB</math>, <math>BC</math>, <math>CA</math>, where <math>M</math> is the midpoint of <math>BC</math> | ||
+ | |||
+ | https://youtu.be/MgrghZ2ESAg | ||
Part 3: Prove that <math>\angle AIL + \angle XPM = 90^\circ</math> and <math>\angle AIK + \angle YPM = 90^\circ</math>. | Part 3: Prove that <math>\angle AIL + \angle XPM = 90^\circ</math> and <math>\angle AIK + \angle YPM = 90^\circ</math>. | ||
+ | https://youtu.be/iOp9mnmZyzU | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 01:31, 24 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
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
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)