### 2005 comc#4b problem and (my)solution

the pciture is the 2005 COMC#4

my solution:

Part A is rather obvious and rather ridiculous, if you realize that the lenngth of external tangents are the same.

Part B:

let: BC=a, AC=b,AB=c.

Lemma(well known): CT= s - c, BT=s-b

proof:

let M, N, T be the tangency points of AB, AC, BC with the incircle of the triangle ABC. Then AM = AN, BM = BT, CN = CT. The perimeter is p = 2s = AM + BM + BT + CT + CN + AN = 2AM + 2BM + 2CT = 2AB + 2CT = 2c + 2CT so CT = s - c ,

similarly BT=s-b

the question is asking us to prove:

ABC=r_1 CT+r_2 BT=r_1 (s-c)+r_2 (s-b)=s(r_1+r_2)+(r_1 c+r_2 b)

proof:

OQED = OQGF = s{(r_1 + r_2)/2} so the area of the hexagon ODEQGF is twice this or: s(r_1+r_2).

notice that tri ODA is congruent to tri OKA same goes for (OFB, OKB) same goes for :(QEA,QLA) and (QGC,QLC)

from these congruent triangle we get areas of the pentagons ODABF and QEACG are equal to twice the areas of the triangles OAB and QAC respectively so:

ODABF=r_1 c, QEACG=r_2 b.

finally ABC=ODEQGF - (ODABF + QEACG) = (r_1 + r_2) - (r_1 c + r_2 b)

QED

## 2 Comments:

how'd you get we canadian stuff anyways

who do you think?

owen!:D

OOOHH YEAH

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