Tiggsy

Try looking through the list of trig identities that you should have.

I'll take you through an alternative method, but leave you to fill in the detail.

Referring to the diagram given in the question, start by drawing a line through C and the point of contact of the cube with the plane, (call that point D), onto its point of intersection with the line AB. Call that point E.

Draw a diagonal across the top face of the cube from OC to the point of contact with the plane. Call the point on OC,  F.

You now have two (vertical) similar triangles, COE and within that the smaller CFD.

Let the side length of the cube be h and let k be the length of OE

You should then be able to use the similar triangles to show that 

\(\displaystyle h = \frac{ck}{(c\sqrt{2}+k)}.\)

Moving on to the base triangle AOB, the angles AOE and BOE are both equal to 45 deg making it easy to write down the areas of the triangles AOE and BOE. (Area AOE = aksin(45)/2.)

Now say area AOB = area AOE + area BOE and you should be able to show that   \(\displaystyle k= \frac{ab\sqrt2}{(a+b)}.\)

Finally, substitute that into the expression for h and tidy up.

Let O be the origin of a 3D co-ordinate system with OA, OB and OC lying along the Ox, Oy and Oz axes respectively.

The equation of the ABC plane will be

\(\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.\)

If h is the side length of the cube, then the co-ordinates of the point on the cube furthest from the origin will be

(h, h, h), and this point is to lie on the plane, in which case

\(\displaystyle \frac{h}{a}+\frac{h}{b}+\frac{h}{c}=1,\)

so

\(\displaystyle h\left\{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right\}=1 \\ \displaystyle h = \frac{abc}{ab+bc+ca}.\)

First digits, there are only two even numbers, 6 and 8.

Hi Juriemagic,

I used a method like the one you are asking about, but my numbers were different.

I split the counting into four separate groups, first two digits odd - odd, odd - even, even - even or even - odd.

For odd - odd, there are three possibles for the first digit, 5, 7 or 9.

For the second digit there are again three possibles. There are four odd numbers 3, 5, 7, 9 but one of them has already been chosen, so three left to choose from.

The third digit is fixed and there are three possibles for the fourth digit.

That means that the total number of odd-odds is (3*3*1*3) = 27.

For the odd-even group it would be (3*3*1*2) = 18. The 2 at the end because one of the even numbers has been knocked out by the even second digit.

The other two, even-even and even-odd I'll leave you to work out, but the total of all four is 65.

As to the answer that you've posted, I think that the first digit in each bracket relates to the choices, odd, 6, 8, so 3, 1, 1.

If that's the case though, I don't understand where the 5 in the second position comes from.

The triangles XWZ and XYW (in that order) are similar.

( The angles XWZ and XYW are both equal to the angle between the tangent and the line XW,

XWZ because of the parallel lines, and XYW by the circle property, angle in the opposite segment. 

Angle WXY is common to both triangles.)

So,

XZ/XW = XW/XY,

XZ = XW^2/XY = 6^2/14 = 36/14 = 18/7.

Then,

YZ = 14 -18/7 = 80/7.

The 66/7 mentioned at the bottom of the question refers to an earlier posting. 

A version of this question, it had XW = 8 rather than 6, was posted July(?)/ August 2020.

Consider the terms \(a^{2},ab,b^{2}.\).

Their arithmetic mean will be greater than or equal to their geometric mean, (equality when \(a=b),\)

so

\(\displaystyle \frac{a^{2}+ab+b^{2}}{3} \geq\sqrt[3]{a^{2}.ab.b^{2}}\\ a^{2}+ab+b^{2}\geq3ab \\\sqrt{a^{2}+ab+b^{2}}\geq\sqrt{3}\sqrt{ab}.\)

Repeat for the other two and add together all three.

Equality when a = b = c.

You shouldn't.

I should have said geometric.

First term 1, common ratio 2^(1/3).

For part (a), there is a third value for x, x = 0.

For part (b), f(x) = g(x) when x= 0 and so cannot be included in the interval.

Look at 1 + 2^(1/3) + 2^(2/3) and think arithmetic progression.