If one of the legs of a certain right triangle is three times the other, in what ratio does the altitude on the hypotenuse divide the hypotenuse?
Geometric solution:
Triangle ADB is similar to triangle ABC so BD/AD = BC/AB ...(1)
Triangle BDC is similar to triangle ABC so CD/BD = BC/AB ...(2)
Multiply (1) and (2) together: (BD/AD)*(CD/BD) = (BC/AB)2
or CD/AD = (3/1)2
so CD/AD = 9/1
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Yes that is really 'neat' Alan.
I had a much more long winded approach
I just let the sides be 1 and 3 so the hypotenuse is sqrt(10)
I dropped that altitude from the hypotenuse and let it be h units long.
This cut the hypotenuse in the ratio x to sqrt(10)-x
so
$$\\1^2+h^2=x^x\\
and\\
3^2+h^2=(\sqrt{10}-x)^2$$
solving this simultaneously you get $$x=\frac{1}{\sqrt{10}}$$
so you have the ratio
$$\\\frac{1}{\sqrt{10}}:\sqrt{10}-\frac{1}{\sqrt{10}}\\\\
\frac{1}{\sqrt{10}}:\frac{10-1}{\sqrt{10}}\\\\
1:9$$
Alan's method is neater - This is just an alternative. I specialise in 'LONG' methods.
Although this was not particularly long.