$M$ is the midpoint of the side $AB$ in an equilateral triangle $\triangle ABC.$ The point $D$ on the side $BC$ is such that $BD : DC = 3 : 1.$ On the line passing through $C$ and parallel to $MD$ there is a point $T$ inside the triangle $\triangle ABC$ such that $\angle CTA = 150.$ Find the $\angle MT D.$ (K. Ivanov )
Problem
Source: Tuymaada 2022 Junior P-7
Tags: geometry, angle
StarLex1
12.09.2022 12:30
Quite bashy
let the line pass through C and parallel to MD is $CT'$
$BMD=x$
$\frac{BM}{MT'}=\frac{BD}{DC}$
$\frac{BM}{MT'} = 3$
$MT' = \frac{BM}{3}$
$AT' = \frac{2BM}{3}$
let $BM=t$
Cosine Rule
$MD^2=t^2+\frac{9t^2}{4} - 2*t*\frac{3}{2}*t*\cos(60)$
$MD^2 = \frac{7t^2}{4}$
$MD = \frac{t\sqrt{7}}{2}$
$\frac{BD}{BC} = \frac{MD}{CT'}$
$\frac{3}{4} = \frac{\frac{t\sqrt{7}}{2}}{CT'}$
$CT' = \frac{2t*\sqrt{7}}{3}$
and also we get the relation of sinx
$\frac{t*\sqrt{7}}{2}{\frac{1}{\sin(60)}} = \frac{\frac{3t}{2}}{\sin(x)}$
$\sin(x) =\frac{3\sqrt{3}}{2\sqrt{7}}$
$\cos(x) = \frac{1}{2\sqrt{7}}$
By Ratio lemma
$\frac{T'T}{TC} = \frac{\frac{2t}{3}*\sin(x-30)}{2t*\cos(x)}=\frac{4}{3}$
$T'T = \frac{4}{7}*\frac{2t*\sqrt{7}}{3}$
$TC = \frac{3}{7}*\frac{2t*\sqrt{7}}{3}$
Now we can find TD and TM using cosine rule
we get $TM = t , TD=\frac{t}{2}$
Cosine Rule Again
$\frac{7t^2}{4}=t^2+\frac{t^2}{4}-2*t*\frac{t}{2}*cos(MTD)$
$\frac{t^2}{2} = -t^2*cos(MTD)$
$cos(MTD)= \frac{-1}{2}$
$MTD=120$
M11100111001Y1R
01.07.2023 20:52
It has a solution by angle chasing.
Mahdi_Mashayekhi
03.08.2024 16:48
Let $N$ be midpoint of $BC$ and let $AT$ meet $AB$ at $G$. Let $ATC$ meet $AB$ at $F$. Let the line perpendicular to $BC$ at $B$ meet $AG$ at $P$. Note that $\angle CFB = 30$ so $\angle FCB = 90$ so $FC \parallel BP$. Now since $\frac{GM}{MB} = \frac{1}{3}$ then $\frac{AG}{GB} = \frac{1}{2}$ and since $AF=AB$ we have $\frac{BG}{GF} = \frac{1}{2}$ so $APB$ and $BCF$ are homothetic. Now since $PM \parallel CA \parallel NM$ we have that $N,M,P$ are collinear. Note that $\angle ABP = \angle AFC = \angle ATP$ so $ATBP$ is cyclic and $\angle APB = \angle BCF = 90 = \angle ANB$ so $N$ lies on $ATBP$ and obviously $M$ is the center. Now we have that $\angle TMB = 2\angle TPB$ and $\angle TDB = \angle TDN = 2\angle TCN$ and since $\angle TPB + \angle TCN = 90$ we have that $TDBM$ is cyclic so $\angle DTM = 120$.