## Sunday, November 1, 2015

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

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

Click the diagram below for more details.

1. Let AO meet the circumcircle at N
1)
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

2. See problem 1062.

1. Ajit sir I need help for the following problem
http://www.artofproblemsolving.com/community/u271471h1157160p5494288

2. I've solved the problem using physics.

3. GG Problem 1161.
Let 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
Similarly EC = EH.
Hence AD + CE = DH + EH = DE = DB

4. This solution is based on identifying AOHC as concyclic points and the rest is proved by angle tracing

Let the foot of the altitudes from A,B & C be denoted as P,Q&R respectively. ABP & CBR are 30-60-90 triangles
=> m(PHC)=60 ----------(1)
and hence m(AHC)=m(AOC)=120 => AOHC are concyclic -----(2)
Connect AO & OC and consider the isosceles triangle AOC (Since O is the circumcenter), we have m(OAC)=m(OCA)=30 -------(3)
From (2) & (3), we have m(OHA)=30
From (1), we have m(PHC)=m(RHA)=60 implying DH is the angle bisector of m(RHA) and triangle ADH is a 30-120-30 isosceles triangle (AD=DH), similaly HE=EC-----------(4)

Hence m(HDR)=60 => BDE is equilateral

Now let m(OAB)=m(OBA)=x => m(OBC)=60-x --------(5)
and m(OAH)=30-x and from (2) we have m(OCH)=30-x
Hence m(C) = m(OCA)+m(OCH)+m(HCP)=30+30-x+30=90-x---------(6)
From right triangle BQC, we have m(CBQ)=x----------(7)
=> m(OBH)=60-2x ---------(8)
In the triangle OEB, m(OEB)=60=> m(BOE)=60+x and from (8), we have m(BHO)=60+x

Hence BOH is isosceles => BO=BH

Since BD=BE, m(DBO)=m(HBE)=x (from (7)) and BO=BH, the triangles DBO and EBH are congruent and hence DO=HE, similarly the triangles BDH and BEO are congruent and hence DH=OE ----------(9)

As BDE is equilateral, BD=DE=DH+HE=AD+CE (From (4))

From (9), we have OE=DH

Also AB=BD+DA
=> AB=BE+DH
=> AB=BE+DO+OH
=> AB=(BE+EC)+OH
=> OH=AB-BC

5. Solution 2

Let AH meet BC at X and CH meet AB at Y

Since < BAX = 30, < BHC = 120 = < AOC
Hence AOHC is concyclic.
Since <OAC = 30, <YHD = 30 and so <BDE = 60 and Tr.BDE is equilateral.

Further since <DBO = 90-C = <EBH, Tr.s BDO and BEH are congruent ASA
So DO = HE

Now ADH and CEH are both 30-30-120 Tr.s with AD = DH and CE = HE
So BD = DE = DH + HE = AD + CE

Also OH =DH = DO = AD - HE = AD -CE
Also OH = AD - CE = (AD + BD) -(CE + BE) = AB -BC

Sumith Peiris
Moratuwa
Sri Lanka