<p>I was having trouble solving this ARML problem, maybe you math whizzes could help me. ABCD is a parallelogram. E is on AB and C is on FG so that DEFG is also a parallelogram. AE/EB=1/5. FC/CG=1/3. The area of parallelogram ABCD=17. Find the area of DEFG.</p>
<p>I haven't actually attempted to solve it properly, but the first thing popping up: Why not try vectorising it?</p>
<p>I'm not much of a vector man myself. This problem smells of AIME material, so all of you USAMO people out there, i expect this to be a breeze for you, lets go guys!</p>
<p>bumpin this. come on y'all. Someone please blow me away with their massive math capabilities.</p>
<p>i got the area as 17.47</p>
<p>i was just thinking, I will try to do it again latter. my ans doesn't seem right.</p>
<p>Hmm... They should have equal areas unless I'm mistaken. DEFG seems to be some kind of transformation of ABCD. Plus, triangles EDG and EDC are congruent, and since they share bases and altitudes with the DEFG and ABCD, they have half their areas, which means the parallelograms have the same area.</p>
<p>Here's a better solution I just found on AoPS:
Triangle EDC = half the area of ABCD
Triangle EDC = half the area of DEFG
And that's it!</p>
<p>It beats trying to prove that EDG and EDC are congruent. :)</p>
<p>I believe I remember this question from the real thing. I just assumed, without proving it to myself, that the 1/2 thing worked in this case -- afterwards we were all confused that it was so easy (most of us stared at it for a minute before just writing down the answer).</p>
<p>yes....
yes!</p>
<p>i found a pretty quick solution you still need one? i agree with the guys above that the parallelograms have the same area.</p>
<p>Yeah 8 years later it’s real likely that the OP is still stuck on the SAME ARML PROBLEM.</p>