A Brilliant Orthocenter

Geometry Level 4

Given $A,B,C$ lie on $xOy$ coordinates such that $A$ is ( $13$ , $43$ ), $B$ is( $18$ , $-11\sqrt{14}$ ) , $C$ is( $-36$ , $19\sqrt{2}$ ).Now $H$ ( $a$ , $b+c\sqrt{2}+d\sqrt{14}$ ) is the orthocenter of the given triangle $ABC$ ,where $a,b,c,d$ are integers.What's the value of $a+b+c+d$ ?

The answer is 46.

1 solution

Yu Ou
Jan 11, 2018

Since the distanse from $O$ to $A$ , $B$ or $C$ are the same,which is $\sqrt{2018}$ ,so we know that $O$ is the circumcenter of triangle $ABC$ .

We also know that the centroid $G$ of the triangle $ABC$ is ( $\frac{1}{3}$ $(13+18-36)$ , $\frac{1}{3}$ $(43-11\sqrt{14}+19\sqrt{2})$ ) =(- $\frac{5}{3}$ , $\frac{43-11\sqrt{14}+19\sqrt{2}}{3}$ ) .

Because of The Euler Line,we know O,G,H are collinear and OH=3OG.

That means H is ( $3(\frac{-5}{3}$ ), $3(\frac{43-11\sqrt{14}+19\sqrt{2}}{3}$ ))=( $-5$ , $43-11\sqrt{14}+19\sqrt{2}$ ).

Thus we have $a=-5$ , $b=43$ , $c=19$ , $d=-11$ .So $a+b+c+d=-5+43+19-11=46$ .

A Brilliant Orthocenter

The answer is 46.

1 solution

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