In acute $\triangle ABC,$ let $D$ be the foot of perpendicular from $A$ on $BC$. Consider points $K, L, M$ on segment $AD$ such that $AK= KL= LM= MD$. Suppose the sum of the areas of the shaded region equals the sum of the areas of the unshaded regions in the following picture. Prove that $BD= DC$.
Problem
Source: RMO 2014 West Bengal Problem 1
Tags: geometry, trapezoid, similar triangles, geometry proposed
07.12.2014 17:22
Use the fact that similar triangles have area scaled by the square of the factor of similarity. The sum of the areas of the shaded regions is $S = (1/4-1/16+1-9/16)S_{ABD}+(1/16+9/16-1/4)S_{ACD}=\frac{10}{16}S_{ABD}+\frac{6}{16}S_{ACD}$. The sum of the unshaded areas is $\bar S = \frac{6}{16}S_{ABD}+\frac{10}{16}S_{ACD}$. If you equal them you get $S_{ABD}=S_{ACD}$ which gives $BD=DC$.
07.12.2014 17:30
benimath wrote: Use the fact that similar triangles have area scaled by the square of the factor of similarity. The sum of the areas of the shaded regions is $S = (1/4-1/16+1-9/16)S_{ABD}+(1/16+9/16-1/4)S_{ACD}=\frac{10}{16}S_{ABD}+\frac{6}{16}S_{ACD}$. The sum of the unshaded areas is $\bar S = \frac{6}{16}S_{ABD}+\frac{10}{16}S_{ACD}$. If you equal them you get $S_{ABD}=S_{ACD}$ which gives $BD=DC$. @Bunny da : Are we given that those small triangles are similar? Or any such extra information about those small triangles?
07.12.2014 17:32
Umm, yeah, you're given that those lines are parallel (though that wasn't explicitly stated in the paper).
08.12.2014 17:56
easiest rmo problem ever .
09.12.2014 19:49
yea it was just very easy just had to do apply thales theorem and area of trapezium and triangle. which easily gives $ BD = DC $
10.12.2014 09:25
What is thales theorem ??
10.12.2014 11:06
basic proportionality theorem.
10.12.2014 12:11
Okay yeah. Didn't know that name for it.
14.06.2015 18:36
It is not given that the lines are parallel. -_-
20.06.2015 22:44
I think that you can call them parallel WLOG.
14.11.2015 17:14
how can you do that they are certainly not given parallel
21.11.2015 16:14
If you don't assume that they are parallel, the problem is impossible.
28.09.2018 21:09
U r sure?
13.10.2019 15:54
Prove that they have to be parallel to have alt areas equal.
01.07.2020 08:01
Just use Area of triangle = 1/2 bh and area of trapezium = (a+b)h/2