# 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

[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 BAC plus angle BDC is 180 degrees. 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 plus angle BAC is 180, then angle BAC should be equal to angles BDN and CDM. We can quickly prove that the triangles ABC, APB, and AQC are similar, so angles BAC = AQC = APB. We also see that angles AQC = BQN = APB = 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 EC/EN = 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.

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