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 Circumcenters and Incenters (Posted on 2017-02-19)

Let I be the incenter of ΔABC.

Let A', B', and C' be the circumcenters of ΔIBC, ΔICA. and ΔIAB.

Prove the following:

A) For each X ∈ {A,B,C} ( line XI intersects the circumcircle of
ΔABC again at X' ).

B) Area(ΔA'B'C') / Area(ΔABC) = R / 2r, where R and r are the

 No Solution Yet Submitted by Bractals Rating: 5.0000 (1 votes)

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 Solution Comment 1 of 1
Using usual notation..

(A)  /AB’I = 2/ACI = C    (1)   (angles subtended by arc AI)
/CB’I = 2/CAI = A           (angles subtended by arc CI)

Adding:   /AB’C = C + A = 180o – B, so the angles AB’C and
ABC are supplementary, making the quadrilateral AB’CB cyclic,
and proving that B’ lies on the circumcircle of triangle ABC.
By similar reasoning, A’ and C’ also lie on the circumcircle.

Using (1) in the isosceles triangle IAB’ gives

/AIB’ = 90o – C/2 = A/2 + B/2

But A/2 and B/2 are internal angles of triangle ABI, so /AIB’
must be its external angle, proving that B, I, B’ are collinear,
and that BI crosses the circumcircle at B’.
By similar reasoning, AI and CI cross the circumcircle at A’ & C’

(B)  Let O denote the circumcentre of triangles ABC and A’B’C’.

/A’OC’   = 2 /A’B’C’        (angles subtended by arc A’C’)
= 2(/A’B’B + /C’B’B)
= 2(/A’AB + /C’CB)
= 2(A/2 + C/2)
= 180o – B
Area of triangle A’OC’ = [A’OC’]  = (1/2)R2 sin(180 – B)

= (1/2)R2 sin B

and by similar reasoning:
Area of triangle A’B’C’  = [B’OC’] + [C’OA’] + [A’OB’]

= (1/2)R2(sinA +sinB + sinC)

= R(2RsinA +2RsinB +2RsinC)/4

= R(a + b + c)/4

Since area of triangle ABC = r(a + b + c)/2, it follows that

[A’B’C’]/[ABC] = R/(2r)

 Posted by Harry on 2017-02-23 13:05:02

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