### Two Cubes

Two cubes, each with integral side lengths, have a combined volume equal to the total of the lengths of their edges. How big are the cubes? [If you find a result by 'trial and error' you'll need to prove you have found all possible solutions.]

### Rationals Between...

What fractions can you find between the square roots of 65 and 67?

### Square Mean

Is the mean of the squares of two numbers greater than, or less than, the square of their means?

# Mediant

##### Stage: 4 Challenge Level:

No one solved this problem completely although Andrei Lazanu's solution came close. It is not necessary to redraw the diagram to complete the set of inequalities. Perhaps you could use the same argument for the last part Andrei.

Part of Andrei's solution

I first observed that a/b is the tangent of angle BAC in Figure 1 and c/d the tangent of angle FDB.

I also see that the measure of angle BAC is greater than the measure of angle DAE.

All angles of interest for the problem are smaller than 90 degrees, and in this region the tangent is a monotonic increasing function (it does not decrease in this range), so the order relation between angles is transferred to tangents.

Now I write the tangents of these angles: $${tan(DAE)} = {DE \over AE} = {a \over b}$$ $${tan(BAC)} = {BC \over AC} = {{a+c} \over {b+d}}$$I have now to transpose the relation between the angles to their tangents, and I obtain the inequality $${a \over b} < { {a+c} \over {b+d}}$$ for $${a \over b} < {c \over d}$$