## Monday, April 11, 2011

### Fox 338

This is a little related to the recent ones, but much simpler. Here we'd appreciate a pure geometric solution if one found.

1. Maybe I don't understand. But if tou take a 12x12 square and divide it in a 9x12 rectangle and three 3x4 small rectangles, then the square is divided into 4 similar rectangles.
Generally, if L is the square side and x,y are the small rectangles width and height, then y=L/3 and x=L/4

2. you are right Cesar. what a shame! what was I thinking? let's correct this one way or another.
-8foxes

3. additional information (a<b) is added. Is this enough to make the claim hold?
Sorry for the earlier error.

These are what make us human, right? Come on we are no machines, but flesh and blood.

4. You mean:
if L is the square side then:
a=L/3

if rectangles must be similars:
a/b=(L-b)/L.

3*b²-3*L*b+L^2=0

whose discriminant is:
9*L²-12*L^2=-3*L²<0

5. http://imageshack.us/photo/my-images/834/fox338.png/

jyu

6. correction: ACB, not ACD

jyu

7. Hello Yu,
How can you say that "C is on the outside of the semicircle"? I just couldn't see that.
Thanks.

8. C is 2 units below the centre of the semicircle whilst the radius of the semicircle is less than 2 units. The semicircle does not touch the base of the square and C is a point on the base.

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