 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Packing 'em In (Posted on 2004-03-02)  ABC is an acute-angled triangle with area 1. A rectangle F(PQRS) has its vertices on the sides of the triangle, with P and Q on BC, R on AC, and S on AB. Another rectangle, G(WXYZ), has its vertices on the sides of triangle ASR, with W and X on RS, Y on AS, and Z on AR.

What is the maximum total area of F and G?

 See The Solution Submitted by DJ Rating: 4.2000 (10 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Possible solution | Comment 1 of 13
Through S draw a parallel to AC; let T be its intersection with BC. Through Z draw another parallel to AC; let V be its intersection with RS.

F equals the area of TCRS, and G equals the area of VRYZ. BST and SZV are similar to BAC. Maximizing F+G equals minimizing the sum of the areas of BST, SZV, and ZAY.

Let's call BC=2L so the height of ABC is 1/L. If BT=2H, then the area of BST is H^2/L^2. If SV=2K, then the area of SZV=K^2/L^2. Finally, ZY=2L-2H-2K, so the area of ZAY=(L-H-K)^2/L^2.

Maximizing F+G equals minimizing (H^2+K^2+(L-H-K)^2)/L^2. Differentiating with regard to H and K and equating to zero, shows that this is minimum if H=K=L/3, and in that case, F+G equals 2/3 of the total area.

PS. In the first draft of this text, I had written "parallel to BC", which made no sense, instead of "parallel to AC".
Edited on March 2, 2004, 2:45 pm
 Posted by Federico Kereki on 2004-03-02 14:07:23 Please log in:
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