0
$\begingroup$

Let $p_1,p_2,p_3$ be the altitudes of $\triangle ABC$ from vertices $A,B,C$ respectively, $\Delta$ is the area of the triangle,$R$ is the circumradius of the triangle,then$\frac{\cos A}{p_1}+\frac{\cos B}{p_2}+\frac{\cos C}{p_3}=$
$(A)\frac{1}{R}$
$(B)\frac{a^2+b^2+c^2}{2R}$
$(C)\frac{\Delta}{2R}$
$(D)$none of these


I found $p_1=2R(\cos A+\cos B\cos C),p_2=2R(\cos B+\cos A\cos C),p_3=2R(\cos C+\cos B\cos A)$

$\frac{\cos A}{p_1}+\frac{\cos B}{p_2}+\frac{\cos C}{p_3}=\frac{\cos A}{2R(\cos A+\cos B\cos C)}+\frac{\cos B}{2R(\cos B+\cos A\cos C)}+\frac{\cos C}{2R(\cos C+\cos B\cos A)}$

$\frac{\cos A}{2R(\sin B\sin C)}+\frac{\cos B}{2R(\sin A\sin C)}+\frac{\cos C}{2R(\sin B\sin A)}$

I am stuck here.

$\endgroup$
3
  • $\begingroup$ It will not be option B) by dimensional analysis. $\endgroup$
    – N.S.JOHN
    Commented Apr 22, 2016 at 14:28
  • $\begingroup$ @N.S.JOHN neither C) for the same reason $\endgroup$
    – Jean Marie
    Commented Apr 22, 2016 at 14:43
  • $\begingroup$ @JeanMarie Yes. $\endgroup$
    – N.S.JOHN
    Commented Apr 22, 2016 at 14:44

2 Answers 2

4
$\begingroup$

By dimensional analysis, we know the result has dimension $\text{length}^{-1}$. This immediately rules out choice $B$ and $C$. If this is an exam, I will probably just write down the result as $A$, move on and revisit the problem when I have time.

When one need to really solve this, one can use choice $A$ as an ansatz. One should express everything in terms of circumradius $R$ and look for simplification.

Notice $ap_1 = bp_2 = cp_3 = 2\Delta$, we have

$$\frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3} = \frac{a \cos A + b \cos B + c\cos C}{2\Delta}$$ In terms of circumradius $R$, we have

$$\begin{cases} a &= 2R\sin A,\\ b &= 2R\sin B,\\ c &= 2R\sin C \end{cases} \quad\text{ and }\quad \Delta = \frac12 R^2(\sin(2A) + \sin(2B) + \sin(2C))$$ The expression at hand is

$$\frac{R(2\sin A\cos A + 2\sin B\cos B + 2\sin C\cos C)}{ R^2(\sin(2A) + \sin(2B) + \sin(2C)} = \frac{1}{R}$$

This means choice $A$ is indeed the correct answer.

$\endgroup$
3
$\begingroup$

we have $$h_a=\frac{2A}{a}$$ etc and we get $$\frac{1}{2A}(a\cos(\alpha)+b\cos(\beta)+c\cos(\beta))=\frac{2R}{abc}\left(\frac{a(b^2+c^2-a^2)}{2bc}+\frac{b(a^2+c^2-b^2)}{2ac}+\frac{c(a^2+b^2-c^2)}{2ab}\right)=$$ $$\frac{R\cdot16(a+b-c)(a+b+c)(a+c-b)(b+c-a)}{16abc}=\frac{16R}{(abc)^2}\cdot A^2=\frac{16R}{(abc)^2}\cdot \frac{(abc)^2}{16\cdot R^2}=\frac{1}{R}$$

$\endgroup$
7
  • $\begingroup$ it is the answer $\text{A}$- to much typing for me!! $\endgroup$ Commented Apr 22, 2016 at 14:55
  • $\begingroup$ The simplification is very difficult in your method,it is easier in the second method below.Thank you for helping me. $\endgroup$ Commented Apr 22, 2016 at 15:28
  • $\begingroup$ i think you have not right, my method uses elementary formulas in the triangle geometry $\endgroup$ Commented Apr 22, 2016 at 15:30
  • $\begingroup$ Sir i am sorry but i tried simplifying it by myself and it takes much time and infact i could not even reach to $\frac{1}{R}$ $\endgroup$ Commented Apr 22, 2016 at 15:33
  • $\begingroup$ you must follow my steps $\endgroup$ Commented Apr 22, 2016 at 15:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .