Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Wednesday, August 9, 2017

### Geometry Problem 1340: Triangle, Incenter, Concentric Circles, Isosceles Triangles, Congruence

Labels:
circle,
concentric circles,
congruence,
incenter,
isosceles,
triangle

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Let the Radius of Circle 1 be R and the incircle of Trianlge ABC be r

ReplyDeletedrop _|_lars from O to AB, AC and BC. Denote the points as X, Y & Z

We have OX=OY=OZ=r

Join O to D and consider the right trianlge OXD.

We have XD^2+OX^2 = R^2

=> XD^2 = R^2-r^2 --------------(1)

Similary join OM and consider the right triangle OYM

we have YM^2=R^2-r^2--------------(2)

From (1) and (2) we can say DX = MY and also we have AX = AY (Tangents to the incircle)

=> DA = MA ----------(3)

As HA.AM = EA.AD

=> HA=EA (HAE is isosceles)

Also We can say the chords AM,ED & GF are equal in length

I will include your conclusion DE = FG = HM. Thanks.

DeleteIf tr AEH is isosceles then tr ADM is also isosceles since HEDM forms a cyclic quad. This is what we'll show.

ReplyDeleteLet the tangent point on AB be X and the tangent on AC be Y.

1. tr OXD is congruent to tr OYM since OM = OD (larg radius) and OX=OY (small radius) and they are right triangles. So XD = YM

2. Similarly tr AXO is congruent to AYO by 2 sides (1 common and the radius) and being a right tr. so AX = AY.

3. Add those together AD = AM so tr. ADM is isosceles.

4. As noted from the cyclic quad tr AEH must be also.

This can be repeated for the other 2 sides.

Problem 1340

ReplyDeleteSuppose that a recorded circle tangents to P, Q, R on the sides AC, CB, BA respectively.

Then ER=RD=HP=PM=GQ=QF(OP=OR=OQ). But AR=AP so AH=AE.

Similar BD=BF and CM=CG.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

In Circle 1, ED, HM and FG are equidistant from the Centre O and are hence equal.

ReplyDeleteIf the tangent point on ED is X and on HM is Y then EX = HY and because AX = AY, AH = AE

Similarly BF = BD and CM = CG

Sumith Peiris

Moratuwa

Sri Lanka