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

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

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## Sunday, November 1, 2015

### Geometry Problem 1161: Scalene Triangle, 60 Degrees, Angle, Orthocenter, Circumcenter, Equilateral Triangle, Congruence

Labels:
60 degrees,
circle,
circumcenter,
congruence,
equilateral,
orthocenter,
scalene,
triangle

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Let AO meet the circumcircle at N

ReplyDelete1)

Then < HCB = 30 = < CBN since AN is a diameter. So HC//BN

Also < BAN = 90-C = < BCN = < CBH so BH //NC

So BHCN is a parellelogram

Now Tr. ONC is equilateral so NC = BH = BO = R where R is the circumradius

2) Easily < CAN = 30 = <ACO hence < AOC = 120 = AHC. So AOHC is cyclic and so < AHD = 30.

So < BDE = 60 and hence Tr. BDE is equilateral

3) This implies that Tr.s BDH and BOE are congruent ASA hence DH = OE and so DO = HE

4) Let BD = DE = BE = p and if AH meets BC at M let ME = q and let CH meet AB at S

BS = a/2 since Tr. BCS is 30-60-90. So DP = p - a/2 = DH/2 hence DH = 2p -a and so HE = p - (2p - a) = a - p = DO

Therefore OH = 3p - 2a

Now q = HE / 2 = (a-p)/2 and BM = c/2 = p - q. So c/2 = p - (a-p)/2 and hence 3p = c + a. (1)

So AD + CE = c-p + a-p = p from (1 ) above and is therefore = to BD

5) AB -BC = c-a = (c-p) - (a-p) = AD - CE = a + c -2a = 3p - 2a = OH

Sumith Peiris

Moratuwa

Sri Lanka

See problem 1062.

ReplyDeleteAjit sir I need help for the following problem

Deletehttp://www.artofproblemsolving.com/community/u271471h1157160p5494288

plz reply me at hrishiashudesh@gmail.com

I've solved the problem using physics.

DeleteGG Problem 1161.

ReplyDeleteLet X be the midpoint of AC.

BH = 2OX = 2R cos B = R = BO.

So <BOH = <BHO ......(i).

Extend BH to cut AC at K and circle at L.

Extend BO to cut the circle at M.

Connect CM.

<ABM = 90 - <AMB = 90 - <ACB = 90 – C = <KBC.

(i.e)<DBO = < EBH ....(ii).

Subtract (ii) from (i) to get

<BDE = <BED = 60 and ΔBDE is equilateral.

Note that ΔBDO and ΔBEH are congruent and

ADH is a 30-120-30 triangle.

So AD = DH.

Similarly EC = EH.

Hence AD + CE = DH + EH = DE = DB