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#1 2012-09-27 21:17:53
- Fistfiz
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Interesting square problem (and a way to solve it)
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 23:14:19)
30+2=28 (Mom's identity)
#2 2012-09-27 22:46:44
- anonimnystefy
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Re: Interesting square problem (and a way to solve it)
Hi Fistfiz
Just enter
to get
Last edited by anonimnystefy (2012-09-27 22:48:05)
The limit operator is just an excuse for doing something you know you can't.
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#3 2012-09-28 01:17:35
- bobbym
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Re: Interesting square problem (and a way to solve it)
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.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
#4 2012-09-28 02:18:00
- Fistfiz
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Re: Interesting square problem (and a way to solve it)
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)