Difference between revisions of "1996 IMO Problems/Problem 2"
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\end{align*}</cmath> | \end{align*}</cmath> | ||
Therefore, <cmath>CP=CQ=\frac{AC}{AB} \cdot BP \Longrightarrow \frac{AC}{PC} =\frac{AB}{PB} \quad \blacksquare</cmath> | Therefore, <cmath>CP=CQ=\frac{AC}{AB} \cdot BP \Longrightarrow \frac{AC}{PC} =\frac{AB}{PB} \quad \blacksquare</cmath> | ||
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+ | ~JoeyW | ||
==See Also== | ==See Also== |
Latest revision as of 13:46, 25 October 2024
Problem
Let be a point inside triangle
such that
Let ,
be the incenters of triangles
,
, respectively. Show that
,
,
meet at a point.
Solution
let ,
be the angle bisectors of
and
, respectively. Notice that they coincide with line
and
. Therefore, it suffices to show
are concurrent.
Let
, and
. Notice that by angle bisector theorem, we have
Therefore, it suffices to show
Now, construct . Connect
. We notice that
,
. Therefore
. Therefore, we have
Therefore,
~JoeyW
See Also
1996 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |