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Difference between revisions of "Ceva I.2"

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We want to prove <math>X,Y,B</math> [[collinear]], so we consider from which which direction we want to prove this. We can prove <math>\angle XYC + \angle BYC = 180</math> to do it, but we don't know any angles, so that probably won't be much use. Instead, we can prove <math>CM, DP, XB</math> collinear, since the intersection of <math>CM</math> and <math>DP</math> is <math>Y</math>. So, let's consider [[Ceva's]] (a concurrency related formula) on <math>\triangle BCP</math>.
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We want to prove <math>X,Y,B</math> [[collinear]], so we consider from which which direction we want to prove this. We can prove <math>\angle XYC + \angle BYC = 180</math> to do it, but we don't know any angles, so that probably won't be much use. Instead, we can prove <math>CM, DP, XB</math> collinear, since the intersection of <math>CM</math> and <math>DP</math> is <math>Y</math>. So, let's consider [[Ceva's Theorem]] (a concurrency related formula) on <math>\triangle BCP</math>.
  
 
Let <math>AB = c, AC = b, BC = a</math>. That means <math>AM = MB = \frac{c}{2}</math>. There are a lot of unknowns here, so let further set <math>AP = x, CD = y</math>. We know that
 
Let <math>AB = c, AC = b, BC = a</math>. That means <math>AM = MB = \frac{c}{2}</math>. There are a lot of unknowns here, so let further set <math>AP = x, CD = y</math>. We know that
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Now, if we extend <math>EM</math> through <math>E</math> and intersect the line at <math>C</math> [[parallel]] to <math>AB</math> at point <math>G</math>, we see <math>\triangle GEC \sim \triangle MEA</math>. Thus, <math>\dfrac{GC}{AM} = \dfrac{EC}{AE} = \dfrac{CD}{BD} \implies \dfrac{GC}{\frac{c}{2}} = \dfrac{y}{a-y} \implies GC = \dfrac{\frac{c}{2} \cdot y}{a-y}</math>. Using <math>\triangle GCX \sim \triangle MPX</math>, <math>\dfrac{CX}{XP} = \dfrac{GC}{PM} = \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}</math>. Thus,
 
Now, if we extend <math>EM</math> through <math>E</math> and intersect the line at <math>C</math> [[parallel]] to <math>AB</math> at point <math>G</math>, we see <math>\triangle GEC \sim \triangle MEA</math>. Thus, <math>\dfrac{GC}{AM} = \dfrac{EC}{AE} = \dfrac{CD}{BD} \implies \dfrac{GC}{\frac{c}{2}} = \dfrac{y}{a-y} \implies GC = \dfrac{\frac{c}{2} \cdot y}{a-y}</math>. Using <math>\triangle GCX \sim \triangle MPX</math>, <math>\dfrac{CX}{XP} = \dfrac{GC}{PM} = \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}</math>. Thus,
 
<cmath>\dfrac{PM}{MB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}=1</cmath>
 
<cmath>\dfrac{PM}{MB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}=1</cmath>
 +
 +
<math>\mathcal{QED}</math>

Latest revision as of 15:38, 26 February 2023

Problem

  • Let $M$ be the midpoint of side $AB$ of triangle $ABC$. Points $D$ and $E$ lie on line segments $BC$ and $CA$, respectively, such that $DE$ and $AB$ are parallel. Point $P$ lies on line segment $AM$. Lines $EM$ and $CP$ intersect at $X$ and lines $DP$ and $CM$ meet at $Y$. Prove that $X,Y,B$ are collinear.

Solution

[asy] import olympiad; size(12cm); pair A=origin, B=(12,0), C=(4,8); draw(A--B--C--cycle); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); pair D=(7,5), E=(2.5,5); dot("$D$",D,NE); dot("$E$",E,NW); draw(D--E); path p = A--B; pair M=midpoint(p); dot("$M$",M,S); pair P=(4.5,0); dot("$P$",P,S); path x = E--M; path y = C--P; draw(E--M, yellow+linewidth(1)); draw(C--P, yellow+linewidth(1)); pair[] i = intersectionpoints(x,y); dot("$X$",i[0],W); path g = D--P; path h = C--M; draw(D--P, purple+linewidth(1)); draw(C--M, purple+linewidth(1)); pair[] j = intersectionpoints(g,h); dot("$Y$",j[0],W-dir(20)); draw(i[0]--j[0]--B, black+dashed+linewidth(0.5)); pair G=E+3.7*dir(125); dot("$G$",G,N); draw(E--G, black+dashed+linewidth(0.5)); draw(C--G, black+dashed+linewidth(0.5)); [/asy]

We want to prove $X,Y,B$ collinear, so we consider from which which direction we want to prove this. We can prove $\angle XYC + \angle BYC = 180$ to do it, but we don't know any angles, so that probably won't be much use. Instead, we can prove $CM, DP, XB$ collinear, since the intersection of $CM$ and $DP$ is $Y$. So, let's consider Ceva's Theorem (a concurrency related formula) on $\triangle BCP$.

Let $AB = c, AC = b, BC = a$. That means $AM = MB = \frac{c}{2}$. There are a lot of unknowns here, so let further set $AP = x, CD = y$. We know that \[\dfrac{PM}{MB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{CX}{XP}\] Now, if we extend $EM$ through $E$ and intersect the line at $C$ parallel to $AB$ at point $G$, we see $\triangle GEC \sim \triangle MEA$. Thus, $\dfrac{GC}{AM} = \dfrac{EC}{AE} = \dfrac{CD}{BD} \implies \dfrac{GC}{\frac{c}{2}} = \dfrac{y}{a-y} \implies GC = \dfrac{\frac{c}{2} \cdot y}{a-y}$. Using $\triangle GCX \sim \triangle MPX$, $\dfrac{CX}{XP} = \dfrac{GC}{PM} = \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}$. Thus, \[\dfrac{PM}{MB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}=1\]

$\mathcal{QED}$

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