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Topic review (newest first)

2012-09-28 02:18:00

Hi bobbym, thank you for your answer...

Wow! This is a nice formula! Thank you.

Have some idea about how to show EF=FG just using triangles properties?

i.e. we can show PFC=QFA in fact
(1) PD=QD;
(2) DA=DC;
(3) angle ADC is common
(1-2-3)=>PAD=DCQ, subtracting PFQD from both we get PFC=QFA.

I'm sure it can be shown in similar ways that FG=FE.

2012-09-28 01:17:35

Hi Fistfiz;

The proof can be done using analytical geometry. Using midpoint formulas and the intersection of lines it is possible to label each point in terms of the sides of the square with length n. See the diagram below.

We then use a simple formula for the area of a triangle when the vertices are known that uses determinants. We will use it twice.

The area of the red shaded area is:

Of course the area of the square with sides of length n is n^2. So

the ratio of the BEFG to ABCD is 4 / 15.

2012-09-27 22:46:44

Hi Fistfiz

Just enter

to get

2012-09-27 21:17:53

Hi guys! Yesterday I found a cool problem:

in the square ABCD, Q and P cut AD and DC in half. Calculate the ratio of areas BEFG/ABCD.

Now, despite this is correct, I find it a little complicated, and I think that some calculations could be avoided. For example, could you show that GF=FE without explicitly calculating it as I did? Or, how would you show that BF lies on BD?

In general, how would you solve the whole problem?

edit: maybe a mod could put a spoiler on my solution... i don't know how to do that

Hope you like this :-)

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