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Proving an inequality (Posted on 2017-09-17) Difficulty: 3 of 5
Let ABC be a right-angled triangle with hypotenuse c = AB.
Let Wa and Wb be the lengths of the angle bisectors from A & B, respectively.

Prove that Wa + Wb ≤ 2c*sqrt( 2−√2).

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

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Solution Solution Comment 1 of 1

Outline proof:

The bisector of angle A has length  W(A) = c*cos(A)/cos(A/2)

                                                = c(2cos(A/2) – sec(A/2))

Thus     W’(A) = -c*sin(A/2)(1 + sec2(A/2)/2)

For 0 < A < pi/2, both sin(A/2) and sec(A/2) are increasing,
so W’(A) is a decreasing function of A. i.e. W is concave.

Length of bisector of angle B is        W(B) = W(pi/2 – A),
whose graph is a reflection of W(A) in the line  A = pi/4.
It follows that W(B) is also concave wrt A and that their
sum will therefore be concave, with a maximum point where
they cross at A = pi/4.

Thus:      W(A) + W(B)   <= 2*W(pi/4)
                                    <= 2c*cos(pi/4)/cos(pi/8)
                                    <= 2c*cos(pi/4)/sqrt((1 + cos(pi/4))/2)
                                    <= 2c/sqrt(1 + 1/sqrt(2))
                                    <= 2c*sqrt(2 – sqrt(2))

Now, is there a simpler approach using geometry?



  Posted by Harry on 2017-09-28 11:57:05
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