Difference between revisions of "2014 IMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
+ | ===Solution 1=== | ||
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
+ | import graph; size(10.60000000000002cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -4.740000000000007, xmax = 16.46000000000002, ymin = -7.520000000000004, ymax = 4.140000000000004; /* image dimensions */ | ||
+ | pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); pen qqwuqq = rgb(0.000000000000000,0.3921568627450985,0.000000000000000); | ||
− | + | draw((1.800000000000002,3.640000000000004)--(-0.2200000000000002,-1.200000000000001)--(7.660000000000009,-1.140000000000001)--cycle, zzttqq); | |
+ | draw(arc((7.660000000000009,-1.140000000000001),0.6000000000000009,140.7958863822920,180.4362538499006)--(7.660000000000009,-1.140000000000001)--cycle, red); | ||
+ | draw(arc((-0.2200000000000002,-1.200000000000001),0.6000000000000009,0.4362538499004549,67.34652063545073)--(-0.2200000000000002,-1.200000000000001)--cycle, qqwuqq); | ||
+ | draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-106.1141136177082,-39.20411361770813)--(1.800000000000002,3.640000000000004)--cycle, qqwuqq); | ||
+ | draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-112.6534793645495,-73.01347936454944)--(1.800000000000002,3.640000000000004)--cycle, red); | ||
+ | /* draw figures */ | ||
+ | draw((1.800000000000002,3.640000000000004)--(-0.2200000000000002,-1.200000000000001), zzttqq); | ||
+ | draw((-0.2200000000000002,-1.200000000000001)--(7.660000000000009,-1.140000000000001), zzttqq); | ||
+ | draw((7.660000000000009,-1.140000000000001)--(1.800000000000002,3.640000000000004), zzttqq); | ||
+ | draw(arc((7.660000000000009,-1.140000000000001),0.6000000000000009,140.7958863822920,180.4362538499006), red); | ||
+ | draw(arc((7.660000000000009,-1.140000000000001),0.5000000000000008,140.7958863822920,180.4362538499006), red); | ||
+ | draw(arc((-0.2200000000000002,-1.200000000000001),0.6000000000000009,0.4362538499004549,67.34652063545073), qqwuqq); | ||
+ | draw(arc((-0.2200000000000002,-1.200000000000001),0.5000000000000008,0.4362538499004549,67.34652063545073), qqwuqq); | ||
+ | draw(arc((-0.2200000000000002,-1.200000000000001),0.4000000000000006,0.4362538499004549,67.34652063545073), qqwuqq); | ||
+ | draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-106.1141136177082,-39.20411361770813), qqwuqq); | ||
+ | draw(arc((1.800000000000002,3.640000000000004),0.5000000000000008,-106.1141136177082,-39.20411361770813), qqwuqq); | ||
+ | draw(arc((1.800000000000002,3.640000000000004),0.4000000000000006,-106.1141136177082,-39.20411361770813), qqwuqq); | ||
+ | draw(arc((1.800000000000002,3.640000000000004),0.6000000000000009,-112.6534793645495,-73.01347936454944), red); | ||
+ | draw(arc((1.800000000000002,3.640000000000004),0.5000000000000008,-112.6534793645495,-73.01347936454944), red); | ||
+ | draw((-1.022670636276736,-6.130338243877306)--(7.660000000000009,-1.140000000000001)); | ||
+ | draw((4.740746205921980,-5.986847110107199)--(-0.2200000000000002,-1.200000000000001)); | ||
+ | draw((1.800000000000002,3.640000000000004)--(4.740746205921980,-5.986847110107199)); | ||
+ | draw((1.800000000000002,3.640000000000004)--(-1.022670636276736,-6.130338243877306)); | ||
+ | draw(circle((3.711084749329270,0.0008695880898521494), 4.110415438128883)); | ||
+ | /* dots and labels */ | ||
+ | dot((1.800000000000002,3.640000000000004),dotstyle); | ||
+ | label("$A$", (1.880000000000002,3.760000000000004), NE * labelscalefactor); | ||
+ | dot((-0.2200000000000002,-1.200000000000001),dotstyle); | ||
+ | label("$B$", (-0.1400000000000008,-1.080000000000000), NE * labelscalefactor); | ||
+ | dot((7.660000000000009,-1.140000000000001),dotstyle); | ||
+ | label("$C$", (7.740000000000012,-1.020000000000000), NE * labelscalefactor); | ||
+ | dot((0.3886646818616330,-1.245169121938651),dotstyle); | ||
+ | label("$Q$", (0.4600000000000001,-1.120000000000000), NE * labelscalefactor); | ||
+ | dot((3.270373102960991,-1.173423555053598),dotstyle); | ||
+ | label("$P$", (3.360000000000004,-1.060000000000000), NE * labelscalefactor); | ||
+ | dot((4.740746205921980,-5.986847110107199),dotstyle); | ||
+ | label("$M$", (4.820000000000006,-5.860000000000003), NE * labelscalefactor); | ||
+ | dot((-1.022670636276736,-6.130338243877306),dotstyle); | ||
+ | label("$N$", (-0.9400000000000020,-6.020000000000004), NE * labelscalefactor); | ||
+ | dot((2.709057008802497,-3.985539257126989),dotstyle); | ||
+ | label("$D$", (2.780000000000003,-3.860000000000002), NE * labelscalefactor); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
− | + | We are trying to prove that the intersection of <math>BM</math> and <math>CN</math>, call it point <math>D</math>, is on the circumcircle of triangle <math>ABC</math>. In other words, we are trying to prove <math>\angle {BDC} + \angle {BAC} = 180</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 <math>\angle {BDC} + \angle {BAC} = 180</math>. ''Note: This is circular reasoning.'' If <math>\angle {BDC} + \angle {BAC} = 180</math>, then <math>\angle {BAC}</math> should be equal to <math>\angle {BDN}</math> and <math>\angle {CDM}</math>. We can quickly prove that the triangles <math>ABC</math>, <math>APB</math>, and <math>AQC</math> are similar, so <math>\angle {BAC} = \angle {AQC} = \angle {APB}</math>. We also see that <math>\angle {AQC} = \angle {BQN} = \angle {APB} = \angle {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>\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=== | ||
+ | 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> | ||
+ | Hence, by the cotangent rule on <math>ABC</math>, we have | ||
+ | <cmath>\cot \angle{BAL} = 2\cot \angle{BAC} + \cot \angle{ABC} = \cot \angle{MBC}.</cmath> | ||
+ | Because the period of cotangent is <math>180^\circ</math>, but angles are less than <math>180^\circ</math>, we have <math>\angle{BAL} = \angle{MBC}.</math> | ||
+ | |||
+ | Similarly, we have <math>\angle{LAC} = \angle{NCB}.</math> Hence, if <math>BM</math> and <math>CN</math> intersect at <math>Z</math>, then <math>\angle{BZC} = 180^\circ - \angle{BAC}</math> by the Angle Sum in a Triangle Theorem. Hence, <math>BACZ</math> is cyclic, which is equivalent to the desired result. | ||
+ | |||
+ | --[[User:Suli|Suli]] 23:27, 7 February 2015 (EST) | ||
+ | |||
+ | ===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 | ||
+ | <cmath>\frac{BA}{BL} = \frac{AN}{AC}.</cmath> | ||
+ | But we also have <math>\angle{ABL} = \angle{CAQ}</math>, so triangles <math>ABL</math> and <math>NAC</math> are similar by <math>SAS</math> similarity. In particular, <math>\angle{ANC} = \angle{BAL}</math>. Similarly, <math>\angle{BMA} = \angle{CAL}</math>, so <math>\angle{ANC} + \angle{BMA} = \angle{BAC}</math>. In addition, angle sum in triangle <math>AQP</math> gives <math>\angle{QAP} = 180^\circ - 2\angle{A}</math>. Therefore, if we let lines <math>BM</math> and <math>CN</math> intersect at <math>T</math>, by Angle Sum in quadrilateral <math>AMTN</math> concave <math>\angle{NTM} = 180^\circ + \angle{A}</math>, and so convex <math>\angle{BTC} = 180^\circ - \angle{A}</math>, which is enough to prove that <math>BACT</math> is cyclic. This completes the proof. | ||
+ | |||
+ | <asy> | ||
+ | size(250); | ||
+ | defaultpen(fontsize(8pt)); | ||
+ | |||
+ | pair A = dir(110); | ||
+ | pair B = dir(210); | ||
+ | pair C = dir(330); | ||
+ | pair Pp = rotate(50, A)*B; | ||
+ | pair P = extension(A,Pp,B,C); | ||
+ | pair Qp = rotate(-70, A)*C; | ||
+ | pair Q = extension(A,Qp,B,C); | ||
+ | pair M = rotate(180,P)*A; | ||
+ | pair N = rotate(180,Q)*A; | ||
+ | path c1 = circumcircle(A,B,C); | ||
+ | pair T = IP(B--M,C--N); | ||
+ | pair L = midpoint(B--C); | ||
+ | |||
+ | draw(A--B--C--cycle^^B--Q--A--P^^Q--N--C^^P--M--B^^A--L); | ||
+ | draw(c1); | ||
+ | |||
+ | dot("$A$", A, dir(100)); | ||
+ | dot("$B$", B, dir(-110)); | ||
+ | dot("$C$", C, dir(-40)); | ||
+ | dot("$P$", P, dir(50)); | ||
+ | dot("$Q$", Q, dir(-170)); | ||
+ | dot("$M$", M, dir(-50)); | ||
+ | dot("$N$", N, dir(-140)); | ||
+ | dot("$T$", T,dir(-90)); | ||
+ | dot("$L$", L, dir(-120)); | ||
+ | </asy> | ||
+ | |||
+ | --[[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> | ||
+ | |||
+ | === Solution 5 === | ||
+ | |||
+ | We use barycentric coordinates. Due to the equal angles, <math>AC</math> is tangent to the circumcircle of <math>ABQ</math> and <math>AB</math> is tangent to the circumcircle of the <math>APC.</math> Therefore, we can use power of a point to solve for side ratios. We have <cmath>A=(1,0,0), B=(0,1,0), C=(0,0,1)</cmath> <cmath>P=(0:a^2-c^2:c^2),Q=(0:b^2:a^2-b^2)</cmath> <cmath>M=(-a^2:2a^2-2c^2:2c^2),N=(-a^2:2b^2:2a^2-2b^2)</cmath> | ||
+ | Therefore, <math>D=(-a^2:2b^2:2c^2),</math> as <math>BM</math> and <math>CN</math> are cevians. Note that <math>(x,y,z)</math> lies on the circumcircle iff <math>a^2yz+b^2xz+c^2xy=0.</math> Substituting the values in, we have <cmath>-4a^2b^2c^2+2a^2b^2c^2+2a^2b^2c^2=0,</cmath> so we are done. <math>\blacksquare</math> | ||
+ | |||
+ | === Solution 6 === | ||
+ | Note that the givens immediately imply that <math>\triangle{ABC} \sim \triangle{QAC} \sim \triangle{PBA}</math>, hence <math>\angle{AQP}=\angle{APQ}=\angle{A}</math>. Let <math>D</math> be the midpoint of BC, <math>E</math> be the midpoint of <math>AC</math>, and <math>F</math> the midpoint of <math>AB</math>. By the similar triangles, we have <math>\angle{BAD}=\angle{AQE}=\angle{AMC}</math>. We also have <math>\angle{BAD}=\angle{BPF}=\angle{MNB}</math>, so we find <math>\angle{AMC}=\angle{MNB}</math>. We note that <math>\angle{AMC}+\angle{CMN}=\angle{AMN}=\angle{AQP}=\angle{A}</math>, so <math>\angle{CMN}+\angle{MNB}=A</math>, which gives that <math>\angle{BKC}=180-\angle{A}</math> and we are done. | ||
+ | |||
+ | As an addition, <math>AK</math> is the A-symmedian in <math>\triangle{ABC}</math>. | ||
+ | |||
+ | === Solution 7 === | ||
+ | Let <math>w</math> be the circumcircle of <math>\triangle{ABC}</math>. Let <math>E</math> and <math>F</math> be the intersection of <math>w</math> with <math>AQ</math> and <math>AP</math>, respectively. By basic angle chasing, we have <math>\angle{ABC} = \angle{CBE}</math> and <math>\angle{ACB} = \angle{BCF}</math>. So if the intersection of <math>BE</math> and <math>CF</math> is <math>D</math>, <math>BC</math> bisects <math>AD</math>. And we know that <math>BC</math> bisects <math>AN</math> and <math>AM</math>, that means <math>N</math>, <math>D</math> and <math>M</math> are collinear. Now, we define the point <math>K</math> which is the intersection of <math>BM</math> and <math>w</math>. And let us say <math>X</math> to the intersection of <math>CK</math> and <math>AE</math>. By Pascal Theorem at <math>FCKBEA</math>: | ||
+ | |||
+ | <math>D</math>, <math>X</math> and <math>M</math> are collinear. That means <math>X</math> is on <math>DM</math>, <math>CK</math> and <math>AE</math>. Therefore <math>X = N</math> and this ends the proof because the intersection of <math>CN</math> and <math>BM</math> is the point <math>K</math> which is on <math>w</math>. <math>\blacksquare</math> | ||
+ | |||
+ | ~Ege Saribas | ||
− | |||
− | |||
− | |||
− | |||
− | |||
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 12:00, 19 May 2024
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.
Solution 5
We use barycentric coordinates. Due to the equal angles, is tangent to the circumcircle of
and
is tangent to the circumcircle of the
Therefore, we can use power of a point to solve for side ratios. We have
Therefore,
as
and
are cevians. Note that
lies on the circumcircle iff
Substituting the values in, we have
so we are done.
Solution 6
Note that the givens immediately imply that , hence
. Let
be the midpoint of BC,
be the midpoint of
, and
the midpoint of
. By the similar triangles, we have
. We also have
, so we find
. We note that
, so
, which gives that
and we are done.
As an addition, is the A-symmedian in
.
Solution 7
Let be the circumcircle of
. Let
and
be the intersection of
with
and
, respectively. By basic angle chasing, we have
and
. So if the intersection of
and
is
,
bisects
. And we know that
bisects
and
, that means
,
and
are collinear. Now, we define the point
which is the intersection of
and
. And let us say
to the intersection of
and
. By Pascal Theorem at
:
,
and
are collinear. That means
is on
,
and
. Therefore
and this ends the proof because the intersection of
and
is the point
which is on
.
~Ege Saribas
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 |