Difference between revisions of "2014 IMO Problems/Problem 4"
(→Solution 1) |
|||
Line 2: | Line 2: | ||
Points <math>P</math> and <math>Q</math> lie on side <math>BC</math> of acute-angled <math>\triangle{ABC}</math> so that <math>\angle{PAB}=\angle{BCA}</math> and <math>\angle{CAQ}=\angle{ABC}</math>. Points <math>M</math> and <math>N</math> lie on lines <math>AP</math> and <math>AQ</math>, respectively, such that <math>P</math> is the midpoint of <math>AM</math>, and <math>Q</math> is the midpoint of <math>AN</math>. Prove that lines <math>BM</math> and <math>CN</math> intersect on the circumcircle of <math>\triangle{ABC}</math>. | Points <math>P</math> and <math>Q</math> lie on side <math>BC</math> of acute-angled <math>\triangle{ABC}</math> so that <math>\angle{PAB}=\angle{BCA}</math> and <math>\angle{CAQ}=\angle{ABC}</math>. Points <math>M</math> and <math>N</math> lie on lines <math>AP</math> and <math>AQ</math>, respectively, such that <math>P</math> is the midpoint of <math>AM</math>, and <math>Q</math> is the midpoint of <math>AN</math>. Prove that lines <math>BM</math> and <math>CN</math> intersect on the circumcircle of <math>\triangle{ABC}</math>. | ||
− | ==Solution 1== | + | ==Solution== |
+ | ===Solution 1=== | ||
<asy> | <asy> | ||
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
Line 62: | Line 63: | ||
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. | 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=== |
Let <math>L</math> be the midpoint of <math>BC</math>. Easy angle chasing gives <math>\angle{AQP} = \angle{APQ} = \angle{BAC}</math>. Because <math>P</math> is the midpoint of <math>AM</math>, the cotangent rule applied on triangle <math>MBA</math> gives us | Let <math>L</math> be the midpoint of <math>BC</math>. Easy angle chasing gives <math>\angle{AQP} = \angle{APQ} = \angle{BAC}</math>. Because <math>P</math> is the midpoint of <math>AM</math>, the cotangent rule applied on triangle <math>MBA</math> gives us | ||
<cmath>\cot \angle{MBC} - \cot \angle{ABC} = 2\cot \angle{BAC}.</cmath> | <cmath>\cot \angle{MBC} - \cot \angle{ABC} = 2\cot \angle{BAC}.</cmath> | ||
Line 73: | Line 74: | ||
--[[User:Suli|Suli]] 23:27, 7 February 2015 (EST) | --[[User:Suli|Suli]] 23:27, 7 February 2015 (EST) | ||
− | ==Solution 3== | + | ===Solution 3=== |
Let <math>L</math> be the midpoint of <math>BC</math>. By AA Similarity, triangles <math>BAP</math> and <math>BCA</math> are similar, so <math>\dfrac{BA}{AP} = \dfrac{BC}{CA}</math> and <math>\angle{BPA} = \angle{BAC}</math>. Similarly, <math>\angle{CQA} = \angle{BAC}</math>, and so triangle <math>AQP</math> is isosceles. Thus, <math>AQ = AP</math>, and so <math>\dfrac{BA}{AQ} = \dfrac{BC}{CA}</math>. Dividing both sides by 2, we have <math>\dfrac{BA}{AN} = \dfrac{BL}{AC}</math>, or | Let <math>L</math> be the midpoint of <math>BC</math>. By AA Similarity, triangles <math>BAP</math> and <math>BCA</math> are similar, so <math>\dfrac{BA}{AP} = \dfrac{BC}{CA}</math> and <math>\angle{BPA} = \angle{BAC}</math>. Similarly, <math>\angle{CQA} = \angle{BAC}</math>, and so triangle <math>AQP</math> is isosceles. Thus, <math>AQ = AP</math>, and so <math>\dfrac{BA}{AQ} = \dfrac{BC}{CA}</math>. Dividing both sides by 2, we have <math>\dfrac{BA}{AN} = \dfrac{BL}{AC}</math>, or | ||
<cmath>\frac{BA}{BL} = \frac{AN}{AC}.</cmath> | <cmath>\frac{BA}{BL} = \frac{AN}{AC}.</cmath> | ||
Line 79: | Line 80: | ||
--[[User:Suli|Suli]] 10:38, 8 February 2015 (EST) | --[[User:Suli|Suli]] 10:38, 8 February 2015 (EST) | ||
+ | |||
+ | ===Solution 4=== | ||
+ | Let <math>D_1</math> be the second intersection of <math>NC</math> with the circumcircle of <math>\triangle ABC,</math> and <math>D_2</math> the second intersection of <math>MB</math> with the circumcircle of <math>\triangle ABC.</math> By inscribed angles, the tangent at <math>C</math> is parallel to <math>AN.</math> Let <math>P_{\infty}</math> denote the point at infinity along line <math>AN.</math> Note that <cmath>(A,D_1;B,C)\stackrel{C}{=}(A,N;Q,P_{\infty})=-1.</cmath> So, <math>ABD_1C</math> is harmonic. Similarly, we can find <math>ABD_2C</math> is harmonic. Therefore, <math>D_1=D_2,</math> which means that <math>BM</math> and <math>CN</math> intersect on the circumcircle. <math>\blacksquare</math> | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 20:07, 28 May 2016
Contents
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
Solution 1
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
.
Let the intersection of
and
be point
, and the intersection of
and
be point
.
Let us assume
. Note: This is circular reasoning. If
, then
should be equal to
and
. We can quickly prove that the triangles
,
, and
are similar, so
. We also see that
. 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)
Solution 4
Let be the second intersection of
with the circumcircle of
and
the second intersection of
with the circumcircle of
By inscribed angles, the tangent at
is parallel to
Let
denote the point at infinity along line
Note that
So,
is harmonic. Similarly, we can find
is harmonic. Therefore,
which means that
and
intersect on the circumcircle.
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 |