2014 IMO Problems/Problem 4

Problem

Points $P$ and $Q$ lie on side $BC$ of acute-angled $\triangle{ABC}$ so that $\angle{PAB}=\angle{BCA}$ and $\angle{CAQ}=\angle{ABC}$. Points $M$ and $N$ lie on lines $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$, and $Q$ is the midpoint of $AN$. Prove that lines $BM$ and $CN$ intersect on the circumcircle of $\triangle{ABC}$.

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 $BM$ and $CN$, call it point $D$, is on the circumcircle of triangle $ABC$. In other words, we are trying to prove $\angle {BDC} + \angle {BAC} = 180$. Let the intersection of $BM$ and $AN$ be point $E$, and the intersection of $AM$ and $CN$ be point $F$. Let us assume $\angle {BDC} + \angle {BAC} = 180$. Note: This is circular reasoning. If $\angle {BDC} + \angle {BAC} = 180$, then $\angle {BAC}$ should be equal to $\angle {BDN}$ and $\angle {CDM}$. We can quickly prove that the triangles $ABC$, $APB$, and $AQC$ are similar, so $\angle {BAC} = \angle {AQC} = \angle {APB}$. We also see that $\angle {AQC} = \angle {BQN} = \angle {APB} = \angle {CPF}$. Also because angles $BEQ$ and $NED, MFD$ and $CFP$ are equal, the triangles $BEQ$ and $NED$, $MDF$ and $FCP$ must be two pairs of similar triangles. Therefore we must prove angles $CBM$ and $ANC, AMB$ and $BCN$ are equal. We have angles $BQA = APC = NQC = BPM$. We also have $AQ = QN$, $AP = PM$. Because the triangles $ABP$ and $ACQ$ are similar, we have $\dfrac {EC}{EN} = \dfrac {BF}{FM}$, so triangles $BFM$ and $NEC$ are similar. So the angles $CBM$ and $ANC, BCN$ and $AMB$ are equal and we are done.

Solution 2

Let $L$ be the midpoint of $BC$. Easy angle chasing gives $\angle{AQP} = \angle{APQ} = \angle{BAC}$. Because $P$ is the midpoint of $AM$, the cotangent rule applied on triangle $MBA$ gives us \[\cot \angle{MBC} - \cot \angle{ABC} = 2\cot \angle{BAC}.\] Hence, by the cotangent rule on $ABC$, we have \[\cot \angle{BAL} = 2\cot \angle{BAC} + \cot \angle{ABC} = \cot \angle{MBC}.\] Because the period of cotangent is $180^\circ$, but angles are less than $180^\circ$, we have $\angle{BAL} = \angle{MBC}.$

Similarly, we have $\angle{LAC} = \angle{NCB}.$ Hence, if $BM$ and $CN$ intersect at $Z$, then $\angle{BZC} = 180^\circ - \angle{BAC}$ by the Angle Sum in a Triangle Theorem. Hence, $BACZ$ is cyclic, which is equivalent to the desired result.

--Suli 23:27, 7 February 2015 (EST)

Solution 3

Let $L$ be the midpoint of $BC$. By AA Similarity, triangles $BAP$ and $BCA$ are similar, so $\dfrac{BA}{AP} = \dfrac{BC}{CA}$ and $\angle{BPA} = \angle{BAC}$. Similarly, $\angle{CQA} = \angle{BAC}$, and so triangle $AQP$ is isosceles. Thus, $AQ = AP$, and so $\dfrac{BA}{AQ} = \dfrac{BC}{CA}$. Dividing both sides by 2, we have $\dfrac{BA}{AN} = \dfrac{BL}{AC}$, or \[\frac{BA}{BL} = \frac{AN}{AC}.\] But we also have $\angle{ABL} = \angle{CAQ}$, so triangles $ABL$ and $NAC$ are similar by $SAS$ similarity. In particular, $\angle{ANC} = \angle{BAL}$. Similarly, $\angle{BMA} = \angle{CAL}$, so $\angle{ANC} + \angle{BMA} = \angle{BAC}$. In addition, angle sum in triangle $AQP$ gives $\angle{QAP} = 180^\circ - 2\angle{A}$. Therefore, if we let lines $BM$ and $CN$ intersect at $T$, by Angle Sum in quadrilateral $AMTN$ concave $\angle{NTM} = 180^\circ + \angle{A}$, and so convex $\angle{BTC} = 180^\circ - \angle{A}$, which is enough to prove that $BACT$ 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]

--Suli 10:38, 8 February 2015 (EST)

Solution 4

Let $D_1$ be the second intersection of $NC$ with the circumcircle of $\triangle ABC,$ and $D_2$ the second intersection of $MB$ with the circumcircle of $\triangle ABC.$ By inscribed angles, the tangent at $C$ is parallel to $AN.$ Let $P_{\infty}$ denote the point at infinity along line $AN.$ Note that \[(A,D_1;B,C)\stackrel{C}{=}(A,N;Q,P_{\infty})=-1.\] So, $ABD_1C$ is harmonic. Similarly, we can find $ABD_2C$ is harmonic. Therefore, $D_1=D_2,$ which means that $BM$ and $CN$ intersect on the circumcircle. $\blacksquare$

Solution 5

We use barycentric coordinates. Due to the equal angles, $AC$ is tangent to the circumcircle of $ABQ$ and $AB$ is tangent to the circumcircle of the $APC.$ Therefore, we can use power of a point to solve for side ratios. We have \[A=(1,0,0), B=(0,1,0), C=(0,0,1)\] \[P=(0:a^2-c^2:c^2),Q=(0:b^2:a^2-b^2)\] \[M=(-a^2:2a^2-2c^2:2c^2),N=(-a^2:2b^2:2a^2-2b^2)\] Therefore, $D=(-a^2:2b^2:2c^2),$ as $BM$ and $CN$ are cevians. Note that $(x,y,z)$ lies on the circumcircle iff $a^2yz+b^2xz+c^2xy=0.$ Substituting the values in, we have \[-4a^2b^2c^2+2a^2b^2c^2+2a^2b^2c^2=0,\] so we are done. $\blacksquare$

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
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