Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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## Topic review (newest first)

anonimnystefy
2013-07-07 02:52:18

The thing I dislike about those kinds of ideas are that they can be tough to get. It does come down to luck and experience a lot.

bob bundy
2013-07-07 02:45:27

hi Stefy,

That is a brilliant way to do it.  Short and no complicated trig stuff.  I am in awe.  ( no dazzled-smiley-face available)

Bob

anonimnystefy
2013-07-07 01:44:24

I do not have an exact definition. Roughly, it means that some times you will think of it, sometimes you won't, and it mostly depends on luck, not unlike many other geometry problems.

My solution extends the line AB and names E the foot of the perpendicular from C to that line. Then I used The basic trig equations to get the result.

bob bundy
2013-07-07 01:05:22

hi Stefy,

What does 'quite a bit of inspection' mean exactly.

Bob

anonimnystefy
2013-07-06 23:33:08

Well, I have a solution that takes 2 or 3 lines to write up, but requires quite a bit of inspection.

bob bundy
2013-07-06 23:25:05

hi mukesh,

Here's an outline of a way to prove this.  see diagram below.

There's no right angle to get tanA easily so I used the sine and cosine rules:

and

Put these together to get tanA and simplify.

work on this expression for tanA, making use of the following:

After much simplification you can get this equal to -2tanB, from which the required result follows.

It's a tough one so expect it to take 2 or 3 pages.  If you get stuck post back where you've got to, and I'll compare your answer with mine.

Bob

mukesh
2013-07-06 04:07:32

if the median 'AD' of a triangle 'ABC' is perpendicular to side AB then prove that    'tanA+2tanB=0'

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