Orthocenter of triangle with vertices on the hyperbola is also on the hyperbola

Question

The points $P(p, 1/p), \,Q(q,1/q), \,R(r, 1/r)$ and $S(s, 1/s)$ lie on the curve $xy = 1$.

a) If $PQ || RS$, show that $pq = rs$.

b) Show that $PQ\perp RS$ if and only if $pqrs=-1$.

c) Use part b to conclude that if a triangle is drawn with its vertices on the rectangular hyperbola $xy = 1$, then the altitudes of the triangle intersect at a common point which also lies on the hyperbola (an altitude of a triangle is the perpendicular from a vertex to the opposite side).

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I have done all the showing parts. I drew out a diagram and can see that it does indeed intersect on the hyperbola, but I'm, not sure what to do now. What is the question asking and how do I show it? Is there algebra involved?

Answer 1

Start with the triangle $PQR$. The altitude dropped from $P$ intersects the unit hyperbola on the segment bounded by $Q$ and $R$; call that intersection $S_P$. Since $PS_P$ is perpendicular to $QR$, the relation in part (b) immediately tells you $S_P$'s coordinates.

Now find the intersection $S_Q$ for the altitude dropped from $Q$ and the intersection $S_R$ for the altitude dropped from $R$. Do $S_P$, $S_Q$, and $S_R$ all have the same coordinates? If so, the three altitudes intersect. If you can show they intersect in at most one point, you are done since you have found that point.

Answer 2

Disclaimer: This answer is overkill. Hopefully a useful overkill.

I will use this question to determine an explicit formula of the orthocenter (i.e. the intersection point of the altitudes) of a triangle in cartesian coordinates of $\mathbb R^2$.

Suppose we have a (non-degenerate) triangle with vertices $A=(a_1,a_2), B=(b_1,b_2), C=(c_1, c_2)\in\mathbb R^2$.

Then the line spanned by the altitude starting at $A$ is $$\left\{A+t\begin{pmatrix}c_2-b_2 \\ b_1-c_1\end{pmatrix}:t\in\mathbb R\right\}\subset\mathbb R^2.$$

Analogously, the line spanned by the altitude starting at $B$ is $$\left\{B+s\begin{pmatrix}a_2-c_2 \\ c_1-a_1\end{pmatrix}:s\in\mathbb R\right\}\subset\mathbb R^2.$$

To compute the intersection, we have to solve the following linear equation system: $$A+t\begin{pmatrix}c_2-b_2 \\ b_1-c_1\end{pmatrix}=B+s\begin{pmatrix}a_2-c_2 \\ c_1-a_1\end{pmatrix}$$ for $s,t\in\mathbb R$.

The solution is $$t=\frac{\left(a_1-b_1\right) \left(b_1-c_1\right)+\left(a_2-b_2\right) \left(b_2-c_2\right)}{a_2 \left(b_1-c_1\right)+b_2 \left(c_1-a_1\right)+c_2 \left(a_1-b_1\right)}, s=\frac{\left(a_1-b_1\right) \left(a_1-c_1\right)+\left(a_2-b_2\right) \left(a_2-c_2\right)}{b_2 \left(a_1-c_1\right)+a_2 \left(c_1-b_1\right)+c_2 \left(b_1-a_1\right)}.$$

Plugging in $t$ into the altitude starting at $A$ gives the following explicit formula for the orthocenter:

$$\bbox[15px,border:1px groove navy]{\begin{pmatrix}\frac{\left(c_2-b_2\right) \left(\left(a_1-b_1\right) \left(a_1-c_1\right)+\left(a_2-b_2\right) \left(a_2-c_2\right)\right)}{b_2 \left(a_1-c_1\right)+a_2 \left(c_1-b_1\right)+c_2 \left(b_1-a_1\right)}+a_1\\\frac{\left(b_1-c_1\right) \left(\left(a_1-b_1\right) \left(a_1-c_1\right)+\left(a_2-b_2\right) \left(a_2-c_2\right)\right)}{b_2 \left(a_1-c_1\right)+a_2 \left(c_1-b_1\right)+c_2 \left(b_1-a_1\right)}+a_2\end{pmatrix}.}$$

Now, doing some tedious computations (thanks, Mathematica), you see that for $a_2= \frac1{a_1},b_2=\frac1{b_1}, c_2=\frac1{c_1}$, the above equals $$\begin{pmatrix} -\frac1{a_1 b_1 c_1}\\ -a_1 b_1 c_1 \end{pmatrix}.$$

This point lies on the hyperbola so we are done.