Let $A$, $B$, and $C$ be three points on the curve $xy = 1$ (which is a rectangular hyperbola). Prove that the orthocenter of $\triangle ABC$ also lies on the curve $xy = 1$.
I have given up on this problem...
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Choose $(p,1/p);(q,1/q);(r,1/r)$ to be three coordinates – lab bhattacharjee Aug 06 '16 at 14:58
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Use Analytical Geometry to figure this out. – Frank Aug 06 '16 at 21:01
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Let $A = (p,1/p), ~B = (q,1/q),~ C = (r, 1/r)$ be the three points of the triangle, let $H$ be the orthocenter.
Verify that the slope of $AB$ is exactly $-1/pq$, so the slope of the altitude $h_C$ is $pq$. Similarly, the slope of the altitude $h_A$ is $qr$.
Now you can calculate the position of $H$, by intersecting $h_A$ and $h_C$: We have $$A + x\cdot\begin{pmatrix}1\\qr\end{pmatrix} = H = C + y \cdot \begin{pmatrix}1\\pq\end{pmatrix}$$
for some $x$ and $y$. You can solve this linear equation system for $x$ and $y$ and get that $H = (-1/pqr, -pqr)$. This point clearly lies on your given hyperbola.
Anon
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$h_A$ is the set of all points of the form $A+x\cdot\begin{pmatrix}1\qr\end{pmatrix}$, and similar for $h_C$. Clearly $H$ lies on both altitudes. – Anon Aug 06 '16 at 15:52
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Let we start with a straightforward
Lemma 1. If $A,B$ are two points on a rectangular hyperbola and $ACBD$ is a rectangle with its sides being parallel to the hyperbola axis, the centre $O$ of the hyperbola lies on the $CD$ line.
Such almost obvious fact has an important consequence: let $ABC$ a triangle with its vertices lying on a rectangular hyperbola and let $M_A,M_B,M_C$ be the midpoints of $BC,AC,AB$. The centre $O$ of our hyperbola fulfills $$ \widehat{M_A O M_C} = \widehat{A B C} $$ by the previous lemma, hence
Lemma 2. If $A,B,C$ are three distinct points on a rectangular hyperbola with centre $O$,
$O$ belongs to the nine point circle of $ABC$.
If $H$ is the orthocenter of $ABC$, the midpoints of $HA,HB,HC$ also lie on the nine point circle of $ABC$. It follows that the nine-point circles of $ABC,HAB,HAC,HBC$ are the same circle.
Given three distinct points $A,B,C$ in the plane and two perpendicular directions, by Lemma 1 there is only one rectangular hyperbola through $A,B,C$ with axis parallel to the given directions.
Moreover,
Lemma 3. Given a hyperbola (or ellipse) and two parallel chords $AB,CD$ on it, the centre of the hyperbola (ellipse) lies on the segment joining the midpoint of $AB$ with the midpoint of $CD$. The same apply if we replace a chord with a tangent at some point.
Now it is not difficult to figure that our lemmata proves that the hyperbola fixed by $H,A,B$ is the same as the hyperbola fixed by $A,B,C$, so $H$ lies on the hyperbola fixed by $A,B,C$, as wanted.
We may also exploit the fact that any circumconic through the orthocenter of a triangle is a rectangular hyperbola (see Jerabek's hyperbola).
Jack D'Aurizio
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I am facing some problems. Consider the other diagonal of the rectangle for $AB$, and call it line $l$. We know that the nine point circle (call $n$) and the line $l$ pass through the center of the hyperbola for $ABH$. But this is not enough to conclude that the hyperbola for ABC and ABH are same because the line may intersect the circle $n$ at 2 points (good one and the midpoint of AB). How to show that midpoint of AB is not the center of the hyperbola. One more thing, why do we need the 3rd lemma? – Sep 12 '18 at 14:24