Difference between revisions of "2012 IMO Problems/Problem 5"
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==Problem== | ==Problem== | ||
− | Let <math>ABC</math> be a triangle with <math>\angle BCA=90^{\circ}</math>, and let <math>D</math> be the foot of the altitude from <math>C</math>. Let <math>X</math> be a point in the interior of the segment <math>CD</math>. Let <math>K</math> be the point on the segment <math>AX</math> such that <math>BK=BC</math>. Similarly, let <math>L</math> be the point on the segment <math>BX</math> such that <math>AL=AC</math>. Let <math>M=\overline{AL}\cap \overline{BK}</math>. Prove that <math>MK=ML</math> | + | Let <math>ABC</math> be a triangle with <math>\angle BCA=90^{\circ}</math>, and let <math>D</math> be the foot of the altitude from <math>C</math>. Let <math>X</math> be a point in the interior of the segment <math>CD</math>. Let <math>K</math> be the point on the segment <math>AX</math> such that <math>BK=BC</math>. Similarly, let <math>L</math> be the point on the segment <math>BX</math> such that <math>AL=AC</math>. Let <math>M=\overline{AL}\cap \overline{BK}</math>. Prove that <math>MK=ML</math>. |
==Solution== | ==Solution== |
Revision as of 07:58, 20 November 2023
Problem
Let be a triangle with
, and let
be the foot of the altitude from
. Let
be a point in the interior of the segment
. Let
be the point on the segment
such that
. Similarly, let
be the point on the segment
such that
. Let
. Prove that
.
Solution
Let's draw a circumcircle around triangle (=
), a circle with it's center as
and radius as
(=
),
a circle with its center as
and radius as
(=
).
Since the center of
lies on
, the three circles above are coaxial to line
.
Let Line and Line
collide with
on
(
) and
(
), Respectively. Also let
be the point where
and
intersect.
Then, since , by ceva's theorem, the point
lies on the line
.
Since triangles and
are similar,
, Thus
.
In the same way,
.
Therefore, making
concyclic.
In the same way,
is concyclic.
So , and in the same way
. Therefore, the lines
and
are tangent to
and
, respectively.
Since is on the line
, and the line
is the concentric line of
and
,
, Thus
. Since
is in the middle and
,
we can say triangles
and
are congruent. Therefore,
.
Edit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2012 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |