Intelligent Orthocenter

Geometry Level 4

In a triangle $ABC$ , if the altitude drawn from vertex $A$ meets the circumcircle at a point $P=(1,5)$ and the equation of the side $BC$ is, $x + y + 4 = 0.$ Then the orthocenter , $H$ is $(a,b)$ . Find $|a+b|$ .

The answer is 14.

2 solutions

Maria Kozlowska
Oct 16, 2016

Orthocenter and point of intersection of altitude and circumcircle are always equidistant from the foot of the given altitude. In our case we are looking for a point on perpendicular line to BC: $-x+y-4=0$ . The foot of the altitude from $A$ is at $D(-4,0)$ .

$a=-4-(1-(-4)=-9, b=(0-(5-0)=-5 \Rightarrow H(-9,-5)$

$\Rightarrow a+b=-14 \Rightarrow |a+b|=\boxed{14}$ .

Dhanvanth Balakrishnan
Oct 15, 2016

There's a direct formula for finding the image of a point $(x_1,y_1)$ w.r.t. the line $L : ax+by+c = 0$ which is

$\dfrac{x- x_1}{a} = \dfrac{y-y_1}{b} = \dfrac{-2(ax_1 + by_1 + c)}{a^2+b^2}$

I hope this approach would be helpful since you already know the property that the image of the orthocenter w.r.t. any side of the triangle lies on the circumcircle of the triangle, so there's no need to find the foot of the perpendicular of the point to the line as you've done.

P.S. I loved your handwriting. Such a nice and neat solution. (+1) for this. Kudos!

Tapas Mazumdar - 4 years, 1 month ago

Thanks for the tip.

Dhanvanth Balakrishnan - 4 years, 1 month ago

Intelligent Orthocenter

The answer is 14.

2 solutions

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