Difference between revisions of "Ceva I.2"
(→Solution) |
(→Solution) |
||
| (3 intermediate revisions by the same user not shown) | |||
| Line 30: | Line 30: | ||
</asy> | </asy> | ||
| − | 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>. | + | 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 | ||
<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}</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}</cmath> | ||
| − | 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
be the midpoint of side 
of triangle 
. Points 
and 
lie on line segments 
and 
, respectively, such that 
and are parallel. Point 
lies on line segment 
. Lines 
and 
intersect at 
and lines 
and 
meet at 
. Prove that 
are collinear.
be the midpoint of side 
. Points 
and 
and 
, respectively, such that 
and 
lies on line segment 
. Lines 
intersect at 
and lines 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]](http://latex.artofproblemsolving.com/b/2/8/b280db3eb6f027e2996443087a81257ff50b4ea4.png)
We want to prove 
collinear, so we consider from which which direction we want to prove this. We can prove 
to do it, but we don't know any angles, so that probably won't be much use. Instead, we can prove 
collinear, since the intersection of 
and 
is 
. So, let's consider Ceva's Theorem (a concurrency related formula) on 
.
Let 
. That means 
. There are a lot of unknowns here, so let further set 
. We know that
![\[\dfrac{PM}{MB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{CX}{XP}\]](http://latex.artofproblemsolving.com/2/1/6/2167a68baa6f1cd73357aecbd0580aa4885aec4d.png)
Now, if we extend 
through 
and intersect the line at 
parallel to 
at point 
, we see 
. Thus, 
. Using 
, 
. Thus,
![\[\dfrac{PM}{MB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}=1\]](http://latex.artofproblemsolving.com/4/a/8/4a8abd9b813ce4d2c455d9658fcd41b16ca21b6e.png)
