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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

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 :-)

*Last edited by Fistfiz (2012-09-27 01:14:19)*

30+2=28 (Mom's identity)

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

Hi Fistfiz

Just enter

to get

*Last edited by anonimnystefy (2012-09-27 00:48:05)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,183

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

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.

30+2=28 (Mom's identity)

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