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

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

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## Thursday, February 2, 2017

### Geometry Problem 1312: Triangle, 60 Degrees, Circle, Circumcenter, Orthocenter, Altitude, Distance, Perpendicular, Measurement

Labels:
60 degrees,
altitude,
circle,
circumcenter,
distance,
measurement,
orthocenter,
perpendicular,
triangle

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Problem 1312

ReplyDeleteThe extension of OH intersects AB and BC in K,L respectively .According to the problem 761 the triangle BKL is equilateral .Is <AOC=2<ABC=120, <AHC=90+<ABC/2=120.So

AOHC is cyclic.Draw CPM//KL (M in AB, P in AH).Is <AMC=120=<AOC or AMOHC is

, so AH-HC=AH-HP=AP=12.Draw MN perpendicular AP. Is OD=MN=x, MA=2x and

triangle MAN from pythagorean theorem 4x^2=x^2+36 or x=2√(3 ).

APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

Extend AH to meet BC at E

ReplyDeleteThe perpendicular from O to BC cuts BC in half and say meet at P.

So OPED is a rectangle with PE = OD = x -------(1)

let AB = c and consider the 30-60-90 triangle ABE

we have BE = c/2 and AE = Sqrt(3)c/2 -------(2)

From (1) and (2) EC = PC-PE = BP-PE = BE-PE-PE = c-4x/2 --------(3)

Now consider the 30-60-90 triangle ECH

we have

HE = (c-4x)/2Sqrt(3)

CH = (c-4x)/Sqrt(3) --------(4)

Given AH-CH = 12

=> (AE-HE)-CH = 12

=> Sqrt(3)c/2-(c-4x)/2Sqrt(3)- (c-4x)/Sqrt(3) = 12 (from (2) and (4))

=> 3c-c+8x-2c+4x = 12*2*Sqrt(3)

=> x = 2Sqrt(3)

Let R be the circumradius.

ReplyDeleteSince < AOC = < AHC = 120, AOHC is concyclic.

So < OHD = 30 and OH = 2x

Applying Ptolemy,

2x.b + R. CH = R. AH

But R = b/sqrt3 considering Tr. AOC.

So x = (AH - CH).R/b = (12 / 2)/sqrt3

x = 2.sqrt3

Sumith Peiris

Moratuwa

Sri Lanka

Extend AH to P (P on circle) => ΔHPC equilateral

ReplyDeleteExtend CH to M (M on circle) => ΔABC ≡ ΔAMC so if we draw perpendicular to

CM and AB we get x inradius of AHG (G on AB)

AD = 1/2 AP => AD = DH+HP = DH+HC (1) But AH-HC = 12 (2)

From (1) and (2) DH = 6

From ΔODH and pythagore theorem x = 2√3

https://goo.gl/photos/Mo9Wudh9iSaMpX4S8

ReplyDeleteLet c= AH, a= HC

Let AH cut circle ABC at E

Note that BC is the perpen. Bisector of HE => HCE is equilateral

So HE=HC=a

D is the midpoint of AE so AD= ½(a+c)

AOK is 30-60-90 triangle so AO= AC/sqrt(3)

In triangle AHC we have angle AHC= 120 and AC^2= a^2+c^2+a.c

In triangle AOD => x^2= AO^2-AD^2

X^2= 1/3(a^2+c^2+a.c)-1/4(a+c)^2

After simplification we get 12.x^2= (a-c)^2= 144

So x=2.sqrt(3)

Sumith Peiris, Moratuwa, Sri Lanka

ReplyDelete- This solution is EXCELLENT!

Thank u.

Delete