STEP I, II, III 1999 solutions

Aaah. I see. I will try that now.

In the meantime, these are my solutions for questions 1, 3, 4, 5, and 7 of paper III. Incidentally, one of my interview questions was to prove that there were only five regular polyhedra (last part of question 4), but by considering the sum of angles at each vertex instead... Much less long-winded, I think, even though it would be necessary to assume that all regular polyhedra are convex.

Also, I think this post needs updating... (There's also a thread for 2000 now.)

STEP II, Question 13
Don't even quote me on this question, there is absolutely no guidance within it.

Step I Question 9

When the hare overtakes, both have travelled 0.5km, but the tortoise has travelled for 0.5hours longer. Using $time=\frac{distance}{speed}$ for both:

$\frac{0.5}{v}=\frac{0.5}{V}+0.5 \Rightarrow \frac{1}{v}=\frac{1}{V}+1=\frac{V+1}{V}$

Therefore, $v=\frac{V}{V+1}$...(A)

When the hare passes the tortoise on the way back, the hare has travelled $X+1.25$ whilst the tortoise has travelled $X-1.25$. However, the tortoise has been racing for 1 hour longer.

Using $time=\frac{distance}{speed}$ again:
$\frac{X+1.25}{V}+1=\frac{X-1.25}{v} \Rightarrow \frac{X+1.25+V}{V}=\frac{X-1.25}{v}$

Therefore, $v=\frac{V(X-1.25)}{X+1.25+V}$...(B)

From, (A) and (B)

$\frac{V}{V+1}=\frac{V(X-1.25)}{X+1.25+V} \Rightarrow \frac{1}{V+1}=\frac{X-1.25}{X+1.25+V}$
This gives: $X+1.25+V=V(X-1.25)+X-1.25 \Rightarrow V(X-2.25)=2.5 \Rightarrow V=\frac{2.5}{X-2.5}$. Multiplying top and bottom by 4 gives $V=\frac{10}{4X-9}$ as required.

From A we know that $v=\frac{V}{V+1}$. Substituting expression for V gives:

$v=\frac{\frac{10}{4X-9}}{\frac{10+4X-9}{4x-9}} = \frac{10}{4X+1}$.
So, $v=\frac{10}{4X+1}$.

The hare finishes 1.5hrs. This means she has travelled 2X in 1.5hrs.
So, $\frac{2X}{V}=1.5 \Rightarrow 4X=3V$
Substituting expression for V,
$4X=\frac{30}{4X-9} \Rightarrow 16X^2-36X-30=0 \Rightarrow 8X^2-18x-15=0$.
Solving the quadratic but since $X>0$, $X=\frac{18+\sqrt{804}}{16}=\frac{18+2\sqrt{201}}{16}=\frac{9+\sqrt{201}}{8}$.
So, finally $X=\frac{9+\sqrt{201}}{8}$.

I'm currently struggling my way through II/3, will type up when done.