Question

In a triangle ABC, $a=7,b=5,c=3.p_1,p_2,p_3$ are the altitudes from $A,B,C$.

Then find the last digit of $4(a{p}_1+b{p}_2+c{p}_3{)}^2$

Answer 1

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{\frac{15}{2}\times \frac{1}{2}\times \frac{5}{2}\times \frac{9}{2}}$
$\Delta =\frac{15\sqrt{3}}{4}$
Also $\Delta =\frac{1}{2}a{p}_1=\frac{1}{2}b{p}_2=\frac{1}{2}c{p}_3$
$\Rightarrow a{p}_1=b{p}_2=c{p}_3=2\Delta$
$\therefore (a{p}_1+b{p}_2+c{p}_3{)}^2=(2\Delta +2\Delta +2\Delta {)}^2$
$=36\times \frac{225\times 3}{16}$
$\Rightarrow 4(a{p}_1+b{p}_2+c{p}_3{)}^2=6075$

$\therefore$ the last digit is $5$