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In the regular hexagon ABCDEF, shown in the diagram below, has side length 1, AB⃗=u and BC⃗=v. The point X lies along FC⃗ such that ∣∣∣XC⃗∣∣∣=3∣∣∣FX⃗∣∣∣.

Show that A, X and E are colinear.
[4]
Using a vector method, find the exact value of cos(∠BXA).
[7]

The point M is drawn inside the hexagon along FC⃗.

Given that ∣∣∣BM⃗∣∣∣2+∣∣∣MD⃗∣∣∣2=∣∣∣BD⃗∣∣∣2, find the distance between M and X.
[9]

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Let's solve this hexagon problem step by step. One way to show that A, X and E are colinear, is to find AX⃗ and AE⃗. What can we say about AE⃗ in terms of u and v?

Well, AE⃗=AD⃗+DE⃗. Since it's a regular hexagon, AD⃗=2v and DE⃗=−u, so AE⃗=2v−u?

Perfect! Now, we know that X divides FC⃗ in the ratio 3:1. Can you express FX⃗ in terms of FC⃗?

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Solution to (a)

We are given that in the regular hexagon with side length 1:

AB⃗=uandBC⃗=v.

AE⃗=AD⃗+DE⃗=2v−u

(A)

and

AX⃗=AF⃗+FX⃗

Now we know that X lies along CF such that ∣∣∣XC⃗∣∣∣=3∣∣∣FX⃗∣∣∣, and ∣∣∣FC⃗∣∣∣=∣∣∣FX⃗∣∣∣+∣∣∣XC⃗∣∣∣. Therefore:

∣∣∣FC⃗∣∣∣=4∣∣∣FX⃗∣∣∣

And hence FX⃗=41FC⃗=21u (A). Therefore:

AX⃗=AF⃗+FX⃗=(v−u)+21u=v−21u

(A)

Since we want to show that the points A, X, and E are colinear, it suffices to show that AX⃗ is parallel to AE⃗ (or to XE⃗). Notice that:

AX⃗=v−21u=21(2v−u)=21AE⃗

Since multiplying AE⃗ by the scalar 21 simply scales the vector, we conclude that AX⃗ and AE⃗ have the same direction. Therefore, the points A, X, and E are colinear. R

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