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Difference between revisions of "2014 IMO Problems/Problem 4"

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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 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 (angle <math>BDC</math>) + (angle <math>BAC</math>) = 180. ''Note: This is circular reasoning.'' If angle <math>BDC</math> plus angle <math>BAC</math> is 180, then angle <math>BAC</math> should be equal to angles <math>BDN</math> and <math>CDM</math>. We can quickly prove that the triangles <math>ABC</math>, <math>APB</math>, and <math>AQC</math> are similar, so angles <math>BAC</math> = <math>AQC</math> = <math>APB</math>. We also see that angles <math>AQC = BQN = APB = 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.  
 
Let us assume (angle <math>BDC</math>) + (angle <math>BAC</math>) = 180. ''Note: This is circular reasoning.'' If angle <math>BDC</math> plus angle <math>BAC</math> is 180, then angle <math>BAC</math> should be equal to angles <math>BDN</math> and <math>CDM</math>. We can quickly prove that the triangles <math>ABC</math>, <math>APB</math>, and <math>AQC</math> are similar, so angles <math>BAC</math> = <math>AQC</math> = <math>APB</math>. We also see that angles <math>AQC = BQN = APB = 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>EC/EN = 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==

Revision as of 20:32, 28 February 2016

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 $\dfrac {EC}{EN} = \dfrac {BF}{FM}$, 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.

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

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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