## Thursday, March 17, 2011

### Problem 594: Triangle, Incenter, Incircle, Inradius, Tangency Point, Midpoint, Altitude, Angle, Half the Difference

Geometry Problem
Click the figure below to see the complete problem 594.

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1. Connect BO, per previous problem , we have BF=DO and BFDO is a parallelogram.
So angle(x)=angle(FBO)= angle(ABO)-angle(ABE)
=(180-angle(A)-angle(C))/2-(90-angle(A))
=(angle(A)-angle(C)/2

Peter Tran

2. Here is a trigonometric solution (rather long)

We make use of the following results:

(i)DM = (a - c)/2 (Problem 587)

(ii)r = (s - b) tan B/2

(iii)Cosine Rule

(iv)tan[(A-C)/2]= [(a - c)/(a + c)][cot(B/2)]

Now EM=AM–AE=(b/2)–c cos A

2EM=b–2c cos A

2b.EM=b^2–2bc cos A

= b^2–(b^2+c^2–a^2)=a^2–c^2 using (iii)

b.EM=(a+c).(a–c)/2 = (a+c).DM by (i)

r/FE = OD/FE = DM/EM = b/(a + c)
(Δs ODM,FEM ///)

x = angle FDO = angle EFD (since FE // OD)

tan x = ED/FE = ED. b/[r(a+c)]

=b(EM–DM)/[r(a + c)]

=(2b.EM -2b.DM)/[2r(a +c)]

=[2(a+c).DM -2b.DM]/[2r(a+c)]

=2DM.(a+c–b)/[2r(a+c)]

=(a–c)(a+c–b)/[2r(a+c)]

=(a–c)(s–b)/[r(a+c)]

=[(a–c)/(a+c)][cot(B/2)] by (ii)

= tan[(A-C)/2 ] by (iv)

Hence x=(A–C)/2

Looking forward for a synthetic proof!

Pravin

3. ABE = 90 - A, FBO = x = B/2 - ABE (FBOD parallelogram)
x = B/2 - 90 + A => x = B/2 - 90 + A/2 + A/2
x = 90 - C/2 - 90 + A/2
x = A/2 - C/2

4. By the result of Problem 591,

we have BF = r = OD.

BF, OD being parallel and equal,

BFDO is a paralleogram.

Therefore

x = angle OBF

= angle OBE

= angle OBA - angle EBA

= B/2 - (90deg - A)

= A + B/2 - 90deg

= A/2 + A/2 + B/2 - 90deg

= A/2 - (90deg - A/2 - B/2)

= A/2 - C/2

= (A - C)/2