# 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?

Note that $$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{a}{2A}+\frac{b}{2A}+\frac{c}{2A}=\frac{P}{2A}$$ where $a,b,c$ are the sides of the triangle, $A$ is the area and $P$ is the perimeter.
$$A = \frac{ap}{2}= \frac{bq}{2}= \frac{cr}{2}=a+b+c.$$ $$\frac{ap}{2}=a+b+c$$ $$p=\frac{2(a+b+c)}{a}$$ Similarly, $$q=\frac{2(a+b+c)}{b}$$ $$r=\frac{2(a+b+c)}{c}$$ Now $$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{a}{2(a+b+c)}+\frac{b}{2(a+b+c)}+\frac{c}{2(a+b+c)}$$ $$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{a+b+c}{2(a+b+c)}$$$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{1}{2}$$
$$2(a+b+c)=pa.$$ Thus, $$\sum_{cyc}\frac{1}{p}=\sum_{cyc}\frac{a}{2(a+b+c)}=\frac{1}{2}$$