Difference between revisions of "2014 IMO Problems/Problem 4"
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Let the intersection of <math>BM</math> and <math>AN</math> be point <math>E</math>, and the intersection of <math>AM</math> and <math>CN</math> be point <math>F</math>. | Let the intersection of <math>BM</math> and <math>AN</math> be point <math>E</math>, and the intersection of <math>AM</math> and <math>CN</math> be point <math>F</math>. | ||
Let us assume (angle <math>BDC</math>) + (angle <math>BAC</math>) = 180. ''Note: This is circular reasoning.'' If angle <math>BDC</math> plus angle <math>BAC</math> is 180, then angle <math>BAC</math> should be equal to angles <math>BDN</math> and <math>CDM</math>. We can quickly prove that the triangles <math>ABC</math>, <math>APB</math>, and <math>AQC</math> are similar, so angles <math>BAC</math> = <math>AQC</math> = <math>APB</math>. We also see that angles <math>AQC = BQN = APB = CPF</math>. Also because angles <math>BEQ</math> and <math>NED, MFD</math> and <math>CFP</math> are equal, the triangles <math>BEQ</math> and <math>NED</math>, <math>MDF</math> and <math>FCP</math> must be two pairs of similar triangles. Therefore we must prove angles <math>CBM</math> and <math>ANC, AMB</math> and <math>BCN</math> are equal. | Let us assume (angle <math>BDC</math>) + (angle <math>BAC</math>) = 180. ''Note: This is circular reasoning.'' If angle <math>BDC</math> plus angle <math>BAC</math> is 180, then angle <math>BAC</math> should be equal to angles <math>BDN</math> and <math>CDM</math>. We can quickly prove that the triangles <math>ABC</math>, <math>APB</math>, and <math>AQC</math> are similar, so angles <math>BAC</math> = <math>AQC</math> = <math>APB</math>. We also see that angles <math>AQC = BQN = APB = CPF</math>. Also because angles <math>BEQ</math> and <math>NED, MFD</math> and <math>CFP</math> are equal, the triangles <math>BEQ</math> and <math>NED</math>, <math>MDF</math> and <math>FCP</math> must be two pairs of similar triangles. Therefore we must prove angles <math>CBM</math> and <math>ANC, AMB</math> and <math>BCN</math> are equal. | ||
− | We have angles <math>BQA = APC = NQC = BPM</math>. We also have <math>AQ = QN</math>, <math>AP = PM</math>. Because the triangles <math>ABP</math> and <math>ACQ</math> are similar, we have <math>EC | + | We have angles <math>BQA = APC = NQC = BPM</math>. We also have <math>AQ = QN</math>, <math>AP = PM</math>. Because the triangles <math>ABP</math> and <math>ACQ</math> are similar, we have <math>\dfrac {EC}{EN} = \dfrac {BF}{FM}</math>, so triangles <math>BFM</math> and <math>NEC</math> are similar. So the angles <math>CBM</math> and <math>ANC, BCN</math> and <math>AMB</math> are equal and we are done. |
==Solution 2== | ==Solution 2== |
Revision as of 21:32, 28 February 2016
Problem
Points and
lie on side
of acute-angled
so that
and
. Points
and
lie on lines
and
, respectively, such that
is the midpoint of
, and
is the midpoint of
. Prove that lines
and
intersect on the circumcircle of
.
Solution
We are trying to prove that the intersection of and
, call it point
, is on the circumcircle of triangle
. In other words, we are trying to prove angle
plus angle
is 180 degrees.
Let the intersection of
and
be point
, and the intersection of
and
be point
.
Let us assume (angle
) + (angle
) = 180. Note: This is circular reasoning. If angle
plus angle
is 180, then angle
should be equal to angles
and
. We can quickly prove that the triangles
,
, and
are similar, so angles
=
=
. We also see that angles
. Also because angles
and
and
are equal, the triangles
and
,
and
must be two pairs of similar triangles. Therefore we must prove angles
and
and
are equal.
We have angles
. We also have
,
. Because the triangles
and
are similar, we have
, so triangles
and
are similar. So the angles
and
and
are equal and we are done.
Solution 2
Let be the midpoint of
. Easy angle chasing gives
. Because
is the midpoint of
, the cotangent rule applied on triangle
gives us
Hence, by the cotangent rule on
, we have
Because the period of cotangent is
, but angles are less than
, we have
Similarly, we have Hence, if
and
intersect at
, then
by the Angle Sum in a Triangle Theorem. Hence,
is cyclic, which is equivalent to the desired result.
--Suli 23:27, 7 February 2015 (EST)
Solution 3
Let be the midpoint of
. By AA Similarity, triangles
and
are similar, so
and
. Similarly,
, and so triangle
is isosceles. Thus,
, and so
. Dividing both sides by 2, we have
, or
But we also have
, so triangles
and
are similar by
similarity. In particular,
. Similarly,
, so
. In addition, angle sum in triangle
gives
. Therefore, if we let lines
and
intersect at
, by Angle Sum in quadrilateral
concave
, and so convex
, which is enough to prove that
is cyclic. This completes the proof.
--Suli 10:38, 8 February 2015 (EST)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2014 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |