Geometry Problem
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Thursday, March 17, 2011
Problem 594: Triangle, Incenter, Incircle, Inradius, Tangency Point, Midpoint, Altitude, Angle, Half the Difference
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Connect BO, per previous problem , we have BF=DO and BFDO is a parallelogram.
ReplyDeleteSo angle(x)=angle(FBO)= angle(ABO)-angle(ABE)
=(180-angle(A)-angle(C))/2-(90-angle(A))
=(angle(A)-angle(C)/2
Peter Tran
Here is a trigonometric solution (rather long)
ReplyDeleteWe 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
ABE = 90 - A, FBO = x = B/2 - ABE (FBOD parallelogram)
ReplyDeletex = B/2 - 90 + A => x = B/2 - 90 + A/2 + A/2
x = 90 - C/2 - 90 + A/2
x = A/2 - C/2
By the result of Problem 591,
ReplyDeletewe 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