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{\bf September 2004}

{\bf 5-1 Climbing Solution} \hfill\break Submitter:

Thank you Andrei Lazanu from "Tudor Vianu" National College,
Bucharest, Romania for this solution.

To solve the problem, I plotted three functions for $0 \leq x \leq
\pi/2$. Below are the graphs of the three functions, all three
plotted on the same figure with $y_1={2x\over \pi}$ in blue, $y_2
= \sin x$ in green and $y_3=x$ in red. \par
%Graph Climbing graph.gif here
As seen from the figure, the green graph (i.e. $y = \sin x$) is
situated between the other two graphs in the range of $0 \leq x
\leq \pi/2$. For all $x$ between 0 and $\pi/2$ the point $(x, \sin
x)$ is above the point $(x, 2x/\pi)$ and hence
$${2x\over \pi} \leq \sin x.$$
In this interval the graph of $\sin x$ lies below the line $y=x$
because this line meets the graph of $\sin x$ at the origin and
the gradient of $\sin x$ is less than 1 for all $x$. Hence $\sin x
\leq x$.
\par
For the second part of the problem, I have to prove that:
$${\tan a \over \tan b} < {a \over b}.$$
For any $a$ and $b$ in this interval, both $a$, $b$, $\tan a$ and
$\tan b$ are positive, so that the original inequality is
equivalent to the following one:
$${\tan a \over a} < {\tan b \over b}$$
In the figure below the graph of function $y=\tan x/x$ is plotted
for $0 < x < \pi/2$.
\par %Graph tanx_overx.gif here

I observe that the function is monotonically increasing. This
means that for any $a < b$ (but remaining in the interval $0 < a <
b < \pi/2$), $y(a) < y(b)$ as required.
\par
Alternatively, from the graph of $y = \tan x$ in this interval, if
$a < b$ then the gradient of line joining the origin the point
$(a, \tan a)$ is less than the gradient of the line joining the
origin to the point $(b, \tan b)$ so
$${\tan a \over a} < {\tan b \over b}$$ and hence
$${\tan a \over \tan b} < {a \over b}.$$

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