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Shades of Fermat's Last Theorem

The familiar Pythagorean 3-4-5 triple gives one solution to (x-1)^n + x^n = (x+1)^n so what about other solutions for x an integer and n= 2, 3, 4 or 5?

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Exhaustion

Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2

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Code to Zero

Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135.

Classical Means

Stage: 5 Short Challenge Level: Challenge Level:1

Patrick of Woodbridge School has proved the first part:
$$\eqalign{
AH&=\frac{1}{2}(a+b)\times {\frac{2}{\frac{1}{a}+\frac{1}{b}}}\cr
&=\frac{1}{2}(a+b)\times \frac{2ab}{b+a}\cr
&=ab\cr
&=G^2.}$$
It follows that $H=\frac{G^2}{A}$.

To solve the geometric part there were many different approaches. James of Macmillan Academy first noticed:

A is the radius of the circle. Since the diameter d = a + b the radius $ r =\frac{1}{2}(a+b) = A$. Which is the arithmetic mean.

Rosa from the Phillipines included a nicely labelled diagram to help explain her reasonings:



The approach taken next was to work out G. Saswata from St Xaviers Kolkata India used Pythagoras' theorem:

$$\eqalign{OW &= \frac{1}{2}(a+b) - b = \frac{1}{2}(a-b)\cr
OY &= r = \frac{1}{2}(a+b)}$$

Since OWY is a right angled triangle we can use Pythagoras to say that:
$$\eqalign{
G^2 &= OY^2 - OW^2\cr
&= (\frac{1}{2}(a+b))^2 - (\frac{1}{2}(a-b))^2\cr
&= \frac{1}{4}(a^2 + 2ab + b^2 - a^2 + 2ab - b^2)\cr
&= ab}$$

Hence $G = \sqrt {ab}$ Which is the geometric mean.

Katie from Firrhill Highschool also used Pythagoras and some simultaneous equations to work out H.

The triangle OWY and WZY are both right angled triangles. Using Pythagoras we can write

$$ \eqalign{
WZ^2 &= OW^2- OZ^2 = (\frac{1}{2}(a-b))^2 - (r-H)^2\cr
&= (\frac{1}{2} (a-b))^2 - (\frac{1}{2}(a+b) - H)^2\cr
&= -ab + H(a+b) - H^2 }$$
and
$$\eqalign{
WZ^2 &= G^2 - H^2 = ab - H^2}$$
Hence
$$\eqalign{
ab - H^2 &= -ab +H(a+b) - H^2\cr
H(a+b)&= 2ab\cr
H&= \frac{G^2}{A}}$$

Which we saw in the first part as the Harmonic Mean

Rosa and Patrick took a different approach to finding H. They used the idea of similar triangles.

The triangles OWZ and WZY are similar as the angles are all the same. Therfore the ratio of the two hypotenuses is equal to the ratio of a pair of corresponding sides.


$$\eqalign{
\frac{OY}{WY}&=\frac{WY}{YZ}\cr
\frac{r}{G}&=\frac{G}{H}}$$
Hence
$$\eqalign{
H&= \frac{2G}{r}=\frac {2ab}{a+b}\cr
H&= \frac {2}{\frac{1}{a}+\frac{1}{b}} }$$

Saswata took a trigonemetric approach to finding H:

The angle between $H$ and $G$ is $Y$ say. Then
$$cosY = \frac{H}{G}$$
We can also write
$$cos Y = \frac{G}{r}$$

The calculation continues as above.


It is worth mentioning Zhi's solution from St Marylebone CE School for finding G.

If you imagine taking the mirror image of the diagram so that we now have a complete circle then the diameter and the line 2G become two intersecting chords. We can use the intersecting chord theorem to write
$$G^2 = ab$$

Katie nicely explained why A> G> H:

The hypotenuse is the longest side of a right angled triangle. As $G$ is the hypotenuse of a right angled triangle of which $H$ is another side we can say
$$G> H$$
The diameter is the longest straight line joining two points on a circle. $2A$ is the diameter and $2G$ is another chord, hence
$$A> G$$
Therfore$$A> G> H$$

Everyone who answered part ii also answered part iii using Pythagoras' theorem.

$$\eqalign{
Q^2&= r^2 + (r-b)^2\cr
&= (\frac{1}{2}(a+b))^2+(\frac{1}{2}(a-b))^2\cr
&=\frac{a^2+b^2}{2}}$$
Hence$$Q = \sqrt{\frac{a^2+b^2}{2}}$$