In a triangle $ABC$, points $D$ and $E$ are on segments $BC$ and $AC$ such that $BD=3DC$ and $AE=4EC$. Point $P$ is on line $ED$ such that $D$ is the midpoint of segment $EP$. Lines $AP$ and $BC$ intersect at point $S$. Find the ratio $BS/SD$.
Problem
Source: Indian RMO 2013 Mumbai Region Problem 4
Tags: ratio, geometry unsolved, geometry
Onlygodcanjudgeme
01.02.2014 22:17
$ (1) $ using the Menelaus theorem to the triangle $ ACS $ : $ 1 = \frac {CE}{AC} \cdot \frac {AS}{SP} \cdot \frac {DP}{DE} $ then we take that $ 5 = \frac {AS}{SP} $ $ (2) $ using the Menelaus theorem to the triangle $ AEP $ : $ 1 = \frac {SP}{AP} \cdot \frac {AE}{CE} \cdot \frac {CD}{DS} $ then we take that $ \frac {CD}{DS} = \frac {3}{2} $ $ DC = \frac {BD}{3}, \frac {BS}{SD} = \frac {7}{2} $
SamISI1
04.02.2014 11:07
From theorem Menelaus We have $BS/SD=7/2$
Googlu15
16.07.2017 22:13
I am interested to see a solution using mass points.
biomathematics
16.07.2017 22:56
Well: Let $A = (1,0,0) , B= (0,1,0)$ and $C= (0,0,1)$. Then clearly, $D= (0, \frac{1}{4} , \frac{3}{4}) $ and $E= (\frac{1}{5}, 0 , \frac{4}{5})$. The point $P$ is given by $$ P = 2D-E = 2(0, \frac{1}{4} , \frac{3}{4}) - (\frac{1}{5}, 0 , \frac{4}{5}) = (-\frac{1}{5} , \frac{1}{2} , \frac{7}{10})$$Now $AP$ has equation $7y = 5z$, so the point $S = (0:5:7) = (0 , \frac{5}{12} , \frac{7}{12})$. Since $B,S$ and $D$ are collinear, we have $$\frac{BS}{SD} = |\frac{1 - \frac{5}{12}}{\frac{5}{12} - \frac{1}{4}}| = \frac{7}{2}.$$
Aimingformygoal
28.05.2020 15:47
Just construct a line $ FE$ parallel to $ SD$ and apply thales theorem.
guptaamitu1
28.09.2020 22:16
Let the line through E parallel to AP meet segment BC at K. Since SP is parallel to EK, implies SD/DK = PD/DE = 1. Therefore, SD = DK = 2x (say). so SK = SD + DK = 2x + 2x = 4x. Now since AS is parallel to EK, implies 4x/KC = SK/KC = AE/EC = 4EC/EC = 4. Therefore, KC = x. So DC = DK + KC = 2x + x = 3x. Thus, BD = 3DC = 3(3x) = 9x. Therefore, BS = BD - SD = 9x - 2x = 7x. Hence, BS/SD = 7x/2x = 7/2.
Mikeglicker
28.09.2020 23:59
Onlygodcanjudgeme wrote: $ (1) $ using the Menelaus theorem to the triangle $ ACS $ : $ 1 = \frac {CE}{AC} \cdot \frac {AS}{SP} \cdot \frac {DP}{DE} $ then we take that $ 5 = \frac {AS}{SP} $ $ (2) $ using the Menelaus theorem to the triangle $ AEP $ : $ 1 = \frac {SP}{AP} \cdot \frac {AE}{CE} \cdot \frac {CD}{DS} $ then we take that $ \frac {CD}{DS} = \frac {3}{2} $ $ DC = \frac {BD}{3}, \frac {BS}{SD} = \frac {7}{2} $ Sorry for my ignorance but what do you mean by mass? Are you talking about barycentric coordinates?