2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

$$2(a+b+c)=pa.$$ Thus, $$\sum_{cyc}\frac{1}{p}=\sum_{cyc}\frac{a}{2(a+b+c)}=\frac{1}{2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.