Difference between revisions of "Ceva's Theorem/Problems"
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| − | + | == Introductory == | |
| + | |||
| + | ===Problem=== | ||
| + | |||
| + | Suppose <math>AB, AC</math>, and <math>BC</math> have lengths <math>13, 14</math>, and <math>15</math>, respectively. If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>, find <math>BD</math> and <math>DC</math>. | ||
| + | ===Solution=== | ||
| + | |||
| + | If <math>BD = x</math> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>. From this, we find <math>x = 12</math> and <math>y = 3</math>. | ||
| + | |||
| + | == Intermediate == | ||
| + | |||
| + | === Problem === | ||
| + | |||
| + | Let <math>M</math> be the midpoint of side <math>AB</math> of triangle <math>ABC</math>. Points <math>D</math> and <math>E</math> lie on line segments <math>BC</math> and <math>CA</math>, respectively, such that <math>DE</math> and <math>AB</math> are parallel. Point <math>P</math> lies on line segment <math>AM</math>. Lines <math>EM</math> and <math>CP</math> intersect at <math>X</math> and lines <math>DP</math> and <math>CM</math> meet at <math>Y</math>. Prove that <math>X,Y,B</math> 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 <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 | ||
| + | <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, | ||
| + | <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> | ||
| + | |||
| + | |||
| + | ''[[Ceva's Theorem|Back to main article]]'' | ||
Latest revision as of 21:06, 28 April 2025
Introductory
Problem
Suppose 
, and 
have lengths 
, and 
, respectively. If 
and 
, find 
and 
.
Solution
If 
and 
, then 
, and 
. From this, we find 
and 
.
Intermediate
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.
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)
