Problem 3

Let

ABC

be a triangle. Let

M

and

N

be the points in which the median and the angle bisector, respectively, at

A

meet the side

BC

. Let

Q

and

P

be the points in which the perpendicular at

N

NA

meets

MA

and

BA

, respectively, and

O

the point in which the perpendicular at

P

BA

meets

AN

produced. Prove that

QO

is perpendicular to

BC

Step 2 of 3: Compute the coordinates of P, Q, and O in terms of n

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In plain words

Because $PQ$ is vertical ( $x = n$ ), the $y$ -coordinates of $P$ and $Q$ are just the slopes of $AB$ and $AM$ multiplied by $n$ , and every coordinate scales linearly with $n$ .

N = (n, 0) \implies P = (n, mn), \quad Q = \left(n, \frac{m(b-c)n}{b+c}\right), \quad O = (n(m^2+1), 0)

Detailed analysis

Let $N = (n, 0)$ . The line through $N$ perpendicular to $AN$ (the $x$ -axis) is the vertical line $x = n$ . Intersecting $x = n$ with $AB$ ( $y = mx$ ) and $AM$ ( $y = \frac{m(b-c)}{b+c}x$ ) yields $P = (n, mn)$ and $Q = \left(n, \frac{m(b-c)n}{b+c}\right)$ . In the right triangle $\triangle APO$ with altitude $PN$ to the hypotenuse $AO$ , the metric relation $AN \cdot AO = AP^2$ gives $n \cdot x_O = n^2 + m^2 n^2 = n^2(m^2 + 1)$ , so $O = (n(m^2 + 1), 0)$ .