Geometry Problem

Click the figure below to see the complete problem 599.

## Saturday, April 30, 2011

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## Saturday, April 30, 2011

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Geometry Problem 599: Regular Hexagon, Square, Midpoint, Angle, Degrees

Online Geometry theorems, problems, solutions, and related topics.

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Drop BN perpendicular to AC

ReplyDeletew.l.o.g. let each side of the hexagon be 2

Easy to see

BN = 1,

AN = √3,

AC = 2√3

AM = AH/2 = (AC + CH)/2 = (2√3 + 2)/2 = √3 + 1

NM = AM – AN = 1

∆BNM is right angled isosceles,∠BMN = 45°

x = 135°

it is a lot easier!!!! if you see <AHB=15° you can remember the special triangle 30-15-135.

ReplyDeletemedian traced to the largest side forma angle 45° and x=135°.

Let be O the center of square. Note that OM || AD and BC=OM. Thus triangles CMO and MCB are congruents. We know angle OCH = 45º, thus x = angle MCB=135º.

ReplyDeleteProblem 599

ReplyDeleteForm the equilateral triangle BHK (points M and C are inside the triangle BHK).Then <BKH=60=2.30=2.<BAH so K is circumcenter of triangle ABH.So AK=BK=BH=HK and

MK=MH=MA then triangle BMK=triangle BMH.Therefore < MBH=<MBK=30.Then

<BMA=30+15=45. Therefore <BMH=135.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Problem 599 solution 2

ReplyDeleteForm the equilateral triangle BHK (points Β is inside the triangle BHK).Then

Triangle ABK=triangle ABH then <AKB=<AHB=15 and BK=BH,<BKH=<BHK=45

so <KBH=90=<KMH.Then KBMH is cyclic.So <MBH=<MKH=30.Therefore

<BMH=135.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Drop a perpendicular BP to ACH. Obviously PM= a/2 where a = side of the

ReplyDeleteregular hexagon and the square. But BP also = a/2 considering the 30-60-90

triangle BCP.

Hence BP = PM and so x = 135

Sumith Peiris

Moratuwa

Sri Lanka