same area and perimeter 
The perimeter and the area of a given triangle have the same numerical value when measured via a certain unit of measure. If the lengths of the three altitudes of the triangle are p, q, and r, what is the numerical value of $1/q + 1/p + 1/r$?

So first I set up the three sides' lengths as $a, b, c$ respectively, so that the perimeter and area both would be $a+b+c$. but the area is also equal to $ap/2 + bq/2 + cr/2 = a+b+c$, which leads to $ap + bq+cr = 2a + 2b + 2c$. However, $1/q + 1/p + 1/r = (qr + pr + pq)/(pqr)$, which doesn't seem to help much. How can I solve this?

$$A = \frac{ap}{2}= \frac{bq}{2}= \frac{cr}{2}=a+b+c.$$
$$\frac{ap}{2}=a+b+c$$
$$p=\frac{2(a+b+c)}{a}$$
Similarly,
$$q=\frac{2(a+b+c)}{b}$$
$$r=\frac{2(a+b+c)}{c}$$
Now
$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{a}{2(a+b+c)}+\frac{b}{2(a+b+c)}+\frac{c}{2(a+b+c)}$$$$$$
$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{a+b+c}{2(a+b+c)}$$$$$$$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{1}{2}$$
