Difference between revisions of "2012 IMO Problems/Problem 5"
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==Solution== | ==Solution== | ||
− | + | Let's draw a circumcircle around triangle <math>ABC</math> (= <math>\Gamma</math>), a circle with it's center as <math>A</math> and radius as <math>AC</math> (= <math>\Gamma'</math>), | |
− | a circle with | + | a circle with its center as <math>B</math> and radius as <math>BC</math> (= <math>\Gamma''</math>). |
− | Since the center of | + | Since the center of <math>\Gamma</math> lies on <math>BC</math>, the three circles above are coaxial to line <math>CD</math>. |
− | Let | + | |
− | Then since angle AYB = angle AZB = 90, by ceva's theorem in the | + | |
+ | Let Line <math>AX</math> and Line <math>BX</math> collide with <math>\Gamma</math> on <math>P</math> (<math>\neq A</math>) and <math>Q</math> (<math>\neq B</math>), Respectively. Also let <math>R</math> be the point where <math>AQ</math> and <math>BP</math> intersect. | ||
+ | |||
+ | |||
+ | Then, since <math>\angle AYB = \angle AZB = 90</math>, by ceva's theorem, the point <math>R</math> lies on the line <math>CD</math>. | ||
+ | |||
+ | Since triangles <math>ABC</math> and <math>ACD</math> are similar, <math>AL^2 = AC^2 = AD \cdot AB</math>, Thus <math>\angle ALD = \angle ABL</math>. | ||
+ | In the same way, <math>\angle BKD = \angle BAK</math>. | ||
+ | |||
+ | Therefore, <math>\angle ARD = \angle ABQ = \angle ALD</math> making <math>(A, R, L, D)</math> concyclic. | ||
+ | In the same way, <math>(B, R, K, D)</math> is concyclic. | ||
+ | |||
+ | |||
+ | So <math>\angle ADR = \angle ALR = 90</math>, and in the same way <math>\angle BKR = 90</math>. Therefore, the lines <math>RK</math> and <math>RL</math> are tangent to <math>\Gamma'</math> and <math>\Gamma''</math>, respectively. | ||
+ | |||
+ | |||
+ | Since <math>R</math> is on the line <math>CD</math>, and the line <math>CD</math> is the concentric line of <math>\Gamma'</math> and <math>\Gamma''</math>, <math>RK^2 = RL^2</math>, Thus <math>RK = RL</math>. Since <math>RM</math> is in the middle and <math>\angle ADR = \angle BKR = 90</math>, | ||
+ | we can say triangles <math>RKM</math> and <math>RLM</math> are congruent. Therefore, <math>KM = LM</math>. | ||
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Edit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here. | Edit: I believe that this solution, which was posted on IMO 2012-4's page, was meant to be posted here. | ||
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{{alternate solutions}} | {{alternate solutions}} |
Revision as of 04:12, 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 |