A convex hexagon is called a unit if it has four diagonals of length 1, whose endpoints include all the vertices of the hexagon. Show that there is a unit of area k for any 0 < k ≤ 1. What is the largest possible area for a unit?
Solution
Answer: We can get arbitrarily close to (but not achieve) (3√3)/4 (approx 1.3) by:
To prove the first part, consider the diagram below. Take AB = AC = 1 and angle BAC = 2θ. Take DE = DF = 1 and take the points of intersection X and Y such that AX = DX = AY = DY = 2/3. It is easy to check that the area of the hexagon is sin 2θ. So by taking θ in the interval (0, π/4] we can get any area 0 < k ≤ 1.
It is easy to check that there are six possible configurations for the unit diagonals, as shown in the diagram below.
Consider case 1.
The area of the hexagon is area AEDC + area AFE + area BAC. The part of the segment BF that lies inside AEDC is wasted. The rest goes to provide height for the triangles on bases AE and AC. So area AFE + area BAC can be maximised by taking F close to A and ∠BAC as close to a right angle as possible, so that the height of the triangle BAC (on the base AC) is as large as possible. We can then get arbitrarily close to the area of:
We obviously make AEB a straight line. Now area ADE + area ADC = area ACE + area CDE. So if we regard every point except D as fixed, then we maximise the area by taking ∠EAD = ∠CAD, so that D is the maximum distance from CE. Thus a maximal configuration must have ∠AED = ∠CAD. Similarly, it must have ∠CAD = ∠CAB, so all three angles must be equal. That disposes of case 1.
In cases 2 and 6 we find by a similar (but more tedious argument) the same maximum, although in one case we have to use the argument at the end for the final optimisation. In the other cases the maximum is smaller.
However, all these details would take an already long solution way over length. Does anyone have a better approach?
No. 6 (second case) can be made arbitrarily close to the figure below (with AB = AC = BD = 1). To optimise it, suppose ∠ACB = θ. Area ABDC = area ABC + area BCD. If we fix θ, then BC is fixed, so to maximise area BCD we must take ∠CBD = 90o. But θ cannot be optimal unless also ∠CAD = 90o. We have BA = BD and hence ∠BAD = ∠BDA = 45o - θ/2. Hence 90o = ∠CAD = ∠BAC - ∠BAD = (180o - 2θ) - (45o - θ/2). Hence θ = 30o. So ∠ACD = ∠BDC = 60o and ∠CAB = ∠ABD = 120o. It is easy to check that this has area (3√3)/4.
© John Scholes
jscholes@kalva.demon.co.uk
20 Oct 2000