Difference between revisions of "2014 IMO Problems/Problem 4"
(→Solution 2) |
m (→Solution 3) |
||
(20 intermediate revisions by 5 users not shown) | |||
Line 3: | Line 3: | ||
==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=== | |
− | |||
− | |||
− | |||
− | |||
− | ==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 24: | 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> | ||
− | But we also have <math>\angle{ | + | 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) | --[[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> | ||
{{alternate solutions}} | {{alternate solutions}} |
Latest revision as of 20:03, 1 August 2018
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.
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 |