An ellipse $\Gamma$ and its chord $AB$ are given. Find the locus of orthocenters of triangles $ABC$ inscribed into $\Gamma$.
Problem
Source: Sharygin CR 2019 P21 (Grade 10 - 11)
Tags: geometry
06.03.2019 19:11
I am not too sure about this.... Vrangr wrote: An ellipse $\Gamma$ and its chord $AB$ are given. Find the locus of orthocenters of triangles $ABC$ inscribed into $\Gamma$. Solution . Set $A=(-1,0)$ and $B=(1,0)$ and let $ax^2+by^2+cxy+ey-a=0$ be the equation of $\Gamma,$ define the ellipse $\mathcal E$ as $bx^2+ay^2-cxy-ey-b=0.$ If $AB$ is the major axis the locus is $\boxed{\{(x,y)\mid (x,y)\in\mathcal E,~|x|\ne 1\}}$ otherwise it is $\boxed{\{(x,y)\mid (x,y)\in\mathcal E,~|x|\ne 1\}\cup\{A,B\}}.$ We use cartesian coordinates. Set $A=(-1,0)$, $B=(1,0)$ and $C=(m,n),$ the orthocenter doesn't exist when $(m,n)=(1,0)\text{ or }(-1,0),$ so $ n\ne 0.$ Also let the equation of the given ellipse $\Gamma$ be $ax^2+by^2+cxy+dx+ey+f=0.$ Since $(-1,0)\in\Gamma,$ we have $a-d+f=0$ and since $(1,0)\in\Gamma,$ we have $a+d+f=0.$ Therefore $d=0$ and $f=-a.$ Hence the equation of $\Gamma$ is $ax^2+by^2+cxy+ey-a=0,$ since it is an ellipse, we have $c^2-4ab<0.$ Claim 1. The orthocenter of $\triangle ABC$ is $(m,\tfrac{1-m^2}n).$ Proof. Obviously, as the orthocenter lies on the $C-$altitude of $\triangle ABC,$ the orthocenter has the form $(m,t).$ Let the orthocenter be $H.$ Slope of $AH$ is $\tfrac{t}{m+1}$ and slope of $BC$ is $\tfrac{n}{m-1}.$ Now, as $AH\perp BC,$ we have \[\tfrac{t}{m+1}\cdot \tfrac{n}{m-1}=-1\implies t=\tfrac{1-m^2}{n}.~\square\] Case (i) $|m|\ne 1$ : We want to find the equation of the locus of $H(m,\tfrac{1-m^2}n)$ when $C(m,n)$ moves on $ax^2+by^2+cxy+ey-a=0.$ Hence, we set $x=m$ and $y=\tfrac{1-m^2}n,$ (note that $y\ne 0$ and $x^2\ne 1$) which upon solving gives $m=x$ and $n=\tfrac{1-x^2}y$. Now since $(m,n)$ satisfies $ax^2+by^2+cxy+ey-a=0$ we have, \begin{align*} & ax^2+b(\tfrac{1-x^2}y)^2+cx(\tfrac{1-x^2}y)+e(\tfrac{1-x^2}y)-a=0\\ \implies & \tfrac{x^2-1}{y}(bx^2+ay^2-cxy-ey-b)=0\\ \implies & bx^2+ay^2-cxy-ey-b=0 \end{align*} Therefore, the locus of $H$ has the equation $bx^2+ay^2-cxy-ey-b=0$ but since $|m|\ne 1,$ therefore, points of the form $(1,u)$ and $(-1,u)$ must be omitted. Case (ii) $|m|= 1$ : When $AB$ is the major axis of the given ellipse then the locus is empty set, otherwise, if $C=(1,n)$ or $(-1,n),$ then $ABC$ is right angled. Therefore, the locus in this case are the points $A$ and $B.$ Note that $bx^2+ay^2-cxy-ey-b=0$ is an ellipse as $c^2-4ab<0,$ denote it by $\mathcal E.$ The required locus is $\{(x,y)\mid (x,y)\in\mathcal E,~|x|\ne 1\}\cup\{A,B\}$ if $AB$ is not the major axis and the locus is $\{(x,y)\mid (x,y)\in\mathcal E,~|x|\ne 1\}$ otherwise.$~\blacksquare$
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07.03.2019 01:05
Cause the pencils AH and BH has same cross ratio, the locus is a conic going through A and B And this is limited by two parallel lines so it's an ellipse
04.05.2019 16:36
Wow, this problem is way easier than I expected it to be. I swap $A$ and $C$ in the solution. We claim that the locus is an ellipse through $B$ and $C$. Let $A_1, A_2, A_3, A_4$ be four points on $\Gamma$ distinct from $B$ or $C$, let $H_i$ be the orthocenter of triangle $A_iBC$. \[\measuredangle H_iBH_j = \measuredangle A_iBC - \measuredangle ABH_i - \measuredangle H_jBC = \measuredangle A_iBC + \measuredangle CA_iB - 90^{\circ} - 90^{\circ} + \measuredangle BCA_j = \measuredangle A_iCB + \measuredangle BCA_j = \measuredangle A_iCA_j\]\begin{align*} B(H_1, H_2; H_3, H_4) &= \text{Some product of }\sin\measuredangle H_iBH_j\\ &= \text{Some product of }\sin\measuredangle A_iCA_j\\ &= C(A_1, A_2; A_3, A_4) = B(A_1, A_2; A_3, A_4)\\ &= C(H_1, H_2; H_3, H_4) \end{align*}Therefore, the locus of $H$ is a conic through $B$ and $C$. Note that $H$ is bounded on both sides by two parallel lines, more specifically, the perpendiculars to $BC$ from the two extreme positions of $A$. Hence, it must be an ellipse.
20.09.2021 12:25
Let \(\Omega\) be the circle with diameter \(AB\). Animate \(C\) projectively on \(\Gamma\), and let \(X\) and \(Y\) be the feet from \(A\) and \(B\) to \(\overline{CB}\) and \(\overline{CA}\). By perspectivity at \(B\) and \(A\) respectively, points \(X\) and \(Y\) move projectively along \(\Omega\). Then \(\overline{CA}\) and \(\overline{CB}\) move with degree 1, so the orthocenter \(H=\overline{AX}\cap\overline{BY}\) moves with degree 2. It follows that the locus of orthocenters is a conic, and since this conic is bounded by two parallel lines tangent to \(\Gamma\), the conic must be an ellipse.
24.11.2023 12:05
Note that the map from the $A$-altitude to $B$-altitude is projective by mapping $AD \to D \to \Gamma \to E \to BE$. Since this line between maps is never the identity, the result follows by Steiner inconic.
19.05.2024 00:45
Let the points apart from $A$ and $B$ be $C_1,C_2,C_3,C_4,$ and let the orthocenters of $\triangle ABC_i$ be $H_i.$ Then we simply need to prove that $$\angle H_iBH_j=\angle C_iAC_j,$$as the conic is bounded by two lines that are both perpendicular to $\overline{AB}.$ This is a simple angle chase. \begin{align*} \angle H_iBH_j &= \angle H_1BA+\angle B_2BA \\ &=\angle AC_1H_1+\angle AC_2H_2 \\ &=\angle C_iAC_j.\blacksquare \end{align*}
22.09.2024 19:49
Let $D$ be the reflection of the orthocenter over $AB$. Homography $A,B$ to the circle points. Then we can check that $C$ moves on a circle, and $D$ is the negative inverse of $C$ about the circle whose diameter is the images of the circle points. Thus the locus of $D$ is the inverse of the locus of $C$, which is another circle. Thus in the original problem $D$ lies on a conic. This circle never becomes a line as $C$ moves, since when $C=A,B$ the circle becomes tangent to the ellipse, so the locus of $D$, and thus the orthocenter, is an ellipse.