. In the isosceles triangle $ABC$ with $AC = BC$ we denote by $D$ the foot of the altitude through $C$. The midpoint of $CD$ is denoted by $M$. The line $BM$ intersects $AC$ in $E$. Prove that the length of $AC$ is three times that of $CE$.
Problem
Source: 48th Austrian Mathematical Olympiad Beginners' Competition-2047)
Tags: geometry
PROF65
28.02.2018 03:06
use Menelaus
Kaskade
28.02.2018 03:22
PROF65 wrote: use Menelaus While Menelaus can certainly be used, it seems a bit overkill. It can be proved with simple vectors. We see that $\vec{M}=\frac{1}{2}\vec{C}+\frac{1}{4}\vec{A}+\frac{1}{4}\vec{B}$. Let $\vec{E}=x\vec{A}+\left(1-x\right)\vec{C}$. We need to find $x$, $k$ such that $\vec{E}-\vec{M}=k\left(\vec{M}-\vec{B}\right)$. After solving, this yields $x=\frac{1}{3}$.
whiwho
28.02.2018 21:30
Reflect $M$ through $AB$, call it $D'$ and then is just Tales on $AC:CD'$
parmenides51
10.05.2024 02:05
Let $B’ $ eb the symmetric point of $B$ wrt $C$
Then point $E$ is centroid of triangle $ABB’$ as intersection of two medians $AC$ and line $BM$
Line$ BM$ passes through the midpoint of $AB’$ because of the following lemma in trapezoids
Trapezoid Lemma for 4 collinear points: In any trapezoid, 2 midpoints of bases, intersection point of diagonals and intersection point of non parallel sides are 4 collinear points
YaoAOPS
10.05.2024 02:22
$A = (0, 0), B = (6, 0), C = (3, 6x), D = (3, 0), M = (3, 3x), E = (2, 4x)$, the result is obvious.
Ianis
10.05.2024 04:08
Kaskade wrote: PROF65 wrote: use Menelaus While Menelaus can certainly be used, it seems a bit overkill. It can be proved with simple vectors. We see that $\vec{M}=\frac{1}{2}\vec{C}+\frac{1}{4}\vec{A}+\frac{1}{4}\vec{B}$. Let $\vec{E}=x\vec{A}+\left(1-x\right)\vec{C}$. We need to find $x$, $k$ such that $\vec{E}-\vec{M}=k\left(\vec{M}-\vec{B}\right)$. After solving, this yields $x=\frac{1}{3}$. Vectors are an overkill for a problem that's just Menelaus. But why stop the overkill at vectors when we can destroy it with bary? We have $D=\left (\frac{1}{2},\frac{1}{2},0\right )$, so $M=\left (\frac{1}{4},\frac{1}{4},\frac{1}{2}\right )=(1:1:2)$, and therefore $E=(1:0:2)$, which gives the desired conclusion. Now the elementary solution this problem needs so desperately. We have $[ADM]=[DBM]=[BCM]$, so$$\frac{AE}{EC}=\frac{[AME]}{[CEM]}=\frac{[ABM]}{[BCM]}=2,$$which gives the desired conclusion.