## Saturday, April 30, 2011

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

Geometry Problem
Click the figure below to see the complete problem 599.

1. Drop BN perpendicular to AC
w.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°

2. it is a lot easier!!!! if you see <AHB=15° you can remember the special triangle 30-15-135.
median traced to the largest side forma angle 45° and x=135°.

3. 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.

4. Problem 599
Form 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

5. Problem 599 solution 2
Form 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

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