Let p_1, p_2, p_3 be the altitudes of a trian | vedclass

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(C) We know that the area of triangle $ABC$ is $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$. Thus, $\frac{1}{p_1} = \frac{a}{2

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$\frac{25}{288}$

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(C) We know that the area of triangle $ABC$ is $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$. Thus, $\frac{1}{p_1} = \frac{a}{2\Delta}, \frac{1}{p_2} = \frac{b}{2\Delta}, \frac{1}{p_3} = \frac{c}{2\Delta}$. Also, the ex-radii are given by $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$. From these, $\frac{1}{r_1} = \frac{s-a}{\Delta}, \frac{1}{r_2} = \frac{s-b}{\Delta}, \frac{1}{r_3} = \frac{s-c}{\Delta}$. Summing these, $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{3s - (a+b+c)}{\Delta} = \frac{3s - 2s}{\Delta} = \frac{s}{\Delta} = \frac{1}{r}$. Given $r_1=4, r_2=6, r_3=12$, we have $\frac{1}{r} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} = \frac{3+2+1}{12} = \frac{6}{12} = \frac{1}{2}$, so $r=2$. Now, $\frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = \frac{a^2+b^2+c^2}{4\Delta^2}$. Using the identity $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$ and the relation $\frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}$, we use the formula $\sum \frac{1}{p_1^2} = \frac{1}{4r^2} - \frac{1}{2R^2}$. Alternatively, using $r_1=4, r_2=6, r_3=12$, we find $s=12, \Delta=24, a=6, b=8, c=10$. Then $p_1 = \frac{2\Delta}{a} = 8, p_2 = \frac{2\Delta}{b} = 6, p_3 = \frac{2\Delta}{c} = 4.8$. $\frac{1}{8^2} + \frac{1}{6^2} + \frac{1}{4.8^2} = \frac{1}{64} + \frac{1}{36} + \frac{1}{23.04} = 0.015625 + 0.02777 + 0.0434 = 0.0868 = \frac{25}{288}$.

Let $p_1, p_2, p_3$ be the altitudes of a triangle $ABC$ drawn through the vertices $A, B, C$ respectively. If $r_1=4, r_2=6, r_3=12$ are the ex-radii of triangle $ABC$, then $\frac{1}{p_1^2}+\frac{1}{p_2^2}+\frac{1}{p_3^2}=$

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