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

(Solution 2)
m (Solution 3)
 
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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);
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real labelscalefactor = 0.5; /* changes label-to-point distance */
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pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
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[[File:Screen Shot 2015-01-13 at 4.04.54 PM.png]]
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draw((1.800000000000002,3.640000000000004)--(-0.2200000000000002,-1.200000000000001)--(7.660000000000009,-1.140000000000001)--cycle, zzttqq);
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 +
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</asy>
  
Sorry guys I'm new to AOPS so I don't know how to insert equations and stuff. Please help if you can. Thanks.
+
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.
  
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.
+
===Solution 2===
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. 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 <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 21: Line 71:
  
 
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.
 
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>
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Latest revision as of 20:03, 1 August 2018

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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