In the plane a line $l$ and a point $A$ outside it are given. Find the locus of the incenters of acute-angled triangles having a vertex $A$ and an opposite side lying on $l$.
Problem
Source: Sharygin 2018
Tags: geometry
04.04.2018 22:43
TheDarkPrince wrote: In the plane a line $l$ and a point $A$ outside it are given. Find the locus of the incenters of acute-angled triangles having a vertex $A$ and an opposite side lying on $l$. Let $D$ be the foot of perpendicular from $A$ to $\ell$. Fix $B$ on $\ell$ and let $C$ be the point on $\ell$ with angle $BAC=90^{\circ}$. Varying $B$, obtain the locii of the incenters of $\triangle ABD$ and $\triangle ABC$. The desired locus is the region bounded by these curves. Let $AXDY$ be a square with $XY \parallel \ell$. Then $\overline{DX}$ and $\overline{DY}$ is the locus of the former; while for the latter if $I$ is the incenter of $\triangle ABC$ then $AI=\sqrt{2} d(I, \ell)$ hence the locus is a conic of eccentricity $\sqrt{2}$. Thus, the locus is the region between these two curves.
05.04.2018 09:56
Doubtful about this solution:
05.04.2018 16:14
05.04.2018 16:25
WizardMath wrote:
My solution!!!
29.05.2018 17:09
A bit lengthy... My Solution: Note that the Locus involves a hyperbola. Throughout this solution, we will talk of only one half of the hyperbola (i.e. one arm) which passes through $E,F$ (defined later). Let us allow right or degenerate triangles to avoid confusion. Now let $D$ be the foot of the perpendicular from $A$ to $l$. Now see that for $ABC$ to be non obtuse, $B,C$ must lie on different sides of $D$ in $l$. WoLoG, assume $B$ lies on the right and $C$ on the left. See that coinciding $B,D$ and varying $C$ from $D$ to infinity, incenters lie on a line segment $DE$, where $E$ is a point such that $\angle AED = 90^\circ, AE=ED$. $E$ becomes the incenter when $C$ is at infinity. This is because angle bisector of $\angle ABC$ is a fixed ray, and it has an end point because $\angle BAC \le 90^\circ$. Now see that the locus is symmetrical about $AD$, hence we get segment $DF$, such that $AEDF$ is a square. The cases when $B,C$ have right angles are done. We will work in the coordinate plane for now. Let $I =(x,y)$ be the incenter of $ABC$. Let $D$ be the origin with $l$ the $x-$axis. Let $A = (0,a)$. When $\angle A = 90^\circ$, we get $AI = r\sqrt 2$, where $r$ is the inradius of $ABC$. See that $r = y$, because the distance from $I$ to $BC$ is $r,$ which is $y$. Putting this in the equation, squaring and using distance formula, we get $(y-a)^2 + x^2 = 2y^2$. Hence, we get $x^2 - y^2 - 2ya + a^2 = 0$. So, $(y+a)^2 - x^2 = 2a^2$. So, $\frac{(y+a)^2}{(a \sqrt2)^2} + \frac{x^2}{(a \sqrt{2})^2} = 1$. This is the equation of a hyperbola. Note that this Hyperbola passes through $E,F$ because when $I$ coincides with $E,F$, then $\angle A = 90^\circ$ and one of $B,C$ is $90^\circ$ and the other is $0^\circ$. Let $\mathcal{H}$ be the region enclosed by this Hyperbola and the segments $DE,DF$. We will prove $\mathcal{H}$ is the required Locus. Let $ABC$ be the triangle with incenter $I$. Fixing $B$, we can vary $C$ from $D$ till the extent $\angle BAC = 90^\circ$. See that when $\angle BAC = 90^\circ$, we get $I$ to be on the hyperbola. When $C=D$, we get $I$ on the segment $DF$. See that as $C$ varies uniformly on $l$, $I$ also varies uniformly on a straight line. As the extremes of $I$ happen when $C$ attains the extremes, and noting that both the extremes of $I$ lie on the same side of $AD$, we get $I$ to vary on a line segment entirely on or inside $\mathcal H$. Hence we have proved every incenter lies on $\mathcal H$. We will now prove every point in $\mathcal{H}$ can be an incenter. Let $A'$ be the foot of perpendicular from $I$ to $l$. Note that as $I \in \mathcal{H}$, we have the relation $AI\ge IA' \sqrt{2}$. See that drawing a circle with radius $IA'$ and center $I$, and drawing tangents from $A$ to this circle and intersecting these tangents with $l$ gives us $B,C$. As we know that $IA\ge IA' \sqrt{2}$, we know that $\angle A \le 90^\circ$. The tangents from $A$ to $(I,IA')$ lie on different sides of $AD$. That is because the distance from $I$ to $AD$ is less that the distance $IA'$ because $\angle EDB = 45^\circ = \angle FDC$, and that $I$ lies inside or on $DEF$. So, the circle $(I,IA')$ intersects $AD$ at atleast one point. So, the tangents from $A$ to this circle lie on different sides of $D$. Hence, $B,C$ lie on different sides of $D$ in $l$, $\angle B,\angle C$ must be non obtuse. So, every $I$ inside $\mathcal{H}$ has $B,C$, such that $I$ is the incenter of acute triangle $ABC$. Hence the required locus is $\mathcal{H}$, if we allow right triangles. See that when one angle is right angle, then the incenter lies on the boundary of $\mathcal{H}$. Hence the actual locus is the interior of $\mathcal{H}$.$\blacksquare$.