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== | ||
− | Let | + | Let <math>\Gamma</math>, <math>\Gamma'</math>, <math>\Gamma''</math> be the circumcircle of triangle <math>ABC</math>, the circle with its center as <math>A</math> and radius as <math>AC</math>, and the circle with its center as <math>B</math> and radius as <math>BC</math>, Respectively. |
− | |||
Since the center of <math>\Gamma</math> lies on <math>BC</math>, the three circles above are coaxial to line <math>CD</math>. | Since the center of <math>\Gamma</math> lies on <math>BC</math>, the three circles above are coaxial to line <math>CD</math>. | ||
− | 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 | + | 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 = AQ \cap BP</math>. |
− | Then, since <math>\angle AYB = \angle AZB = 90</math>, by ceva's theorem, | + | Then, since <math>\angle AYB = \angle AZB = 90^{\circ}</math>, by ceva's theorem, <math>R</math> lies on <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>. | 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>. | ||
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− | 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. | + | So <math>\angle ADR = \angle ALR = 90^{\circ}</math>, and in the same way <math>\angle BKR = 90^{\circ}</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 | + | Since <math>R</math> is on <math>CD</math>, and <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^{\circ}</math>, |
we can say triangles <math>RKM</math> and <math>RLM</math> are congruent. Therefore, <math>KM = LM</math>. | we can say triangles <math>RKM</math> and <math>RLM</math> are congruent. Therefore, <math>KM = LM</math>. | ||
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. | ||
+ | |||
+ | ~ Latex edit by Kscv | ||
Latest revision as of 08:30, 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 ,
,
be the circumcircle of triangle
, the circle with its center as
and radius as
, and the circle with its center as
and radius as
, Respectively.
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
.
Then, since , by ceva's theorem,
lies on
.
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
, and
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.
~ Latex edit by Kscv
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 |