I feel like posting iTest solutions.

by pythag011, Sep 16, 2008, 12:01 AM

BTW, how do you solve 96?

I thought that P was the first Brocard point, and did a MASSIVE coordinate bash, and I got it wrong...was it actually the first Brocard point?

Or did I make a computation mistake?

Hm...?

And I made a computation mistake on 98 too....

Outline of solution...

1.Prove that if E is the midpoint of AB, F is the midpoint of BC, G is the midpoint of CD, and H is the midpoint of DA, then I is the intersection of EG and FH. Using Mean Similiarty.

2. Notice that I is the midpoint of EG and teh midpoint of FH by Mean Similarty.

3.Show that m<AH = m< BHG.

4. Now we can easily the points where the incircle is tangent to the sides.

5. TRIG BASH!!!

And I got 9/sqt{101},
which is not right. Maybe I made a computation mistake? Hmm....

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whoa I did all of that for #98 except my teammate solved it with complex #s

also what? mean similarity? no no no USE VECTORS KTHX

by not_trig, Sep 17, 2008, 7:50 PM

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dude no all you like have to realize is that like okay if $ M$, $ N$ are midpoint of $ AB$, $ CD$, then $ MIN$ is a line

then like if incircle is tangent to $ AB$, $ CD$ at $ X$, $ Y$, then you get that $ MX=NY$ and we can compute some good stuff about where the lines are tangent to the incircle

the problem is then a trig bash

by The QuattoMaster 6000, Sep 17, 2008, 10:47 PM

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I was thinking vectors too.

by zephyredx, Sep 18, 2008, 11:58 PM

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