## Wednesday, July 25, 2012

### Problem 791: Regular Hexagon, Midpoints of Side and Diagonal, Equilateral Triangle, Area

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

Click the figure below to see the complete problem 791.

1. http://img822.imageshack.us/img822/2764/problem791.png

Draw lines per attached sketch
Let AB=a
we have AE=BF=BD=a√3
and MD=FN=a/2
∠BDC= ∠AFB=30 and ∠AFN=60
∆BFN congruence to ∆BDM ….. ( case SAS)
So BN=BM and ∠NBM=∠FBD=60 => triangle BMN is equilateral

In right triangle BAN , BN^2=AB^2+AN^2= 7/4 a^2
S(BMN)/S(BFD)= BN^2/BF^2 = (7/4a^2)/(3a^2)= 7/12
But S(ABCDEF)=2S(BFD)
So S(BMN)/S(ABCDEF)=7/24

2. Part 1

Embed the figure in the complex plane,
such that z(D)=1, z(A)=-1.

Let w be the primitive cube root of unity,
i.e. w = cos120 + i sin120

With this notation, we have
z(B)=w, z(F)=w^2, z(C)=-w^2, z(E)=-w

Now
z(M) = 1/2 (1-w^2)
z(N) = 1/2 (-1-w) = 1/2 w^2

Since
z(B) + w z(N) + w^2 z(M)
= w + 1/2 + 1/2 w^2 - 1/2 w
= 1/2 (1 + w + w^2)
= 0

Thus triangle BMN is equilateral.

***
Part 2

Let AB = 2a.

Then
S = 6*sqrt(3)/4 (2a)^2 = 6*sqrt(3) a^2

On the other hand, by cosine law on triangle BCM,
BM^2 = (2a)^2 + a^2 - 2(2a)(a)cos120 = 7 a^2

Area of triangle BMN
= sqrt(3)/4 * BM^2
= 7*sqrt(3)/4 a^2

Hence, Area of BMN = 7/24 S.

3. Using vector approach to solve the problem. Let vectorDC = a, vector CB = b, then vector BA = b - a.

Hence MB = (1/2)a + b ; BN = (1/2)b - (3/2)a ; MN = -a + (3/2)b

For simplicity, let the length of the hexagon be 1, then |a|=|b|=1.

|MB| = |BN| = |MN| = sqrt(7/4)

So triangle BMN is equilateral.

The area of triangle would be 7*sqrt(3)/16.
Yet, the area of hexagon in this case, would be 6*(|aXb|)=3*sqrt(3)/2

q.e.d.

4. WLOG let the side AB of hexagon has the length of 2. Then CM = 1. Since BCM is an interior angle of haxagon, its size is 120 degrees. From cosine law in triangle BCM, we can easily compute that BM = sqrt(7).
Since BE is a diameter of circumcircle, triangle BAE is right. In hexagon, it is also well known that the diameter of circumcircle is twice as long as the side, therefore BE = 4. From Pythagorean Theorem in triangle BAE, we find out that AE = sqrt(12) = 2*sqrt(3), thus AN = sqrt(3). Triangle BAN is also right, we can once more use Pythagorean Theorem and obtain BN = sqrt(7).
By symmetry, CN is perpendicular to AE. Since M is midpoint of CD and CN is paraller to DE, M lies on the perpendicular bisector of EN, triangle MEN is isosceles and MN = ME. But side ME from triangle MED is equal as side BM from triangle CBM, because those two triangles are congruent (case SAS). Thus BM = ME = MN and MN = sqrt(7).
All the sides BM, BN, MN have the length sqrt(7), thus triangle BMN is equilateral.
The area ratio is easily computed by the fact, that regular hexagon can be divided into six equal equilateral triangles with vertex at the center of hexagon and with the length of side 2.