triangles
The coordinates of feet of perpendicular from the vertices
of a triangle on opposite sides are $D(20,25),E(8,16),$
and $F(8,9).$ The number of such triangles are
what i try
we know that point of intersection of feet of perpendicular from vertices to opposite side is orthocenter of triangle.
did not understand what is the use of that definition here
please help me to solve it
Define $K$ as midpoint of $BC$. Call the intersection of $BS$ and $DF$ as $P$. Since $S$ is the center of the circle of $ABC$, we have $\angle BSC=2\angle A$. Since $SB=SC$, SK bisects $\angle BDF$ therefore $\angle BSK=\angle A$. On the other hand since $D$ and $F$ are feet of the altitudes $CAFD$ is cyclic, hence $\angle BDF = \angle A$. Now $\angle A = \angle BSK = \angle BDF$ tells us $KSPD$ is a cyclic quadrilateral. Hence $\angle SKB=\angle DPS=90^{\circ}$.
The orthocenter of $ABC$ is the incenter of the orthic triangle.
If you check that $(10,15)$ is the incenter of the orthic triangle (or simply notice it is the only point lying inside the orthic triangle) it follows that $(5,10),(15,30),(50,-5)$ are the vertices of the original triangle, also since (by computing dot products) the given options form a orthocentric system.
Best Answer
The answer is $4$.
We can say that if $D,E,F$ are three distinct points, then the answer is $4$.
We may suppose the followings :
$D$ is the foot of perpendicular from the vertex $A$ of $\triangle{ABC}$ on $BC$
$E$ is the foot of perpendicular from the vertex $B$ of $\triangle{ABC}$ on $CA$
$F$ is the foot of perpendicular from the vertex $C$ of $\triangle{ABC}$ on $AB$
This answer proves the following three claims :
Claim 1 : The incenter of $\triangle{DEF}$ is the orthocenter of acute $\triangle{ABC}$.
Claim 2 : The excenter of $\triangle{DEF}$ is the orthocenter of obtuse $\triangle{ABC}$.
Claim 3 : If $D,E,F$ are three distinct points, then the number of $\triangle{ABC}$ is $4$.
Claim 1 : The incenter of $\triangle{DEF}$ is the orthocenter of acute $\triangle{ABC}$.
Proof for claim 1 :
Let $H$ be the orthocenter of $\triangle{ABC}$. Since $\triangle{ABD}$ and $\triangle{CBF}$ are similar, we get $\angle{BAD}=\angle{BCF}$. Since $A,F,H,E$ are concyclic, we get $\angle{BAD}=\angle{HEF}$. Also, since $H,D,C,E$ are concyclic, we get $\angle{BCF}=\angle{HED}$. It follows from these that $\angle{HEF}=\angle{HED}$. Similarly, we get $\angle{HFE}=\angle{HFD}$ and $\angle{HDF}=\angle{HDE}$, so the claim follows.$\quad\square$
Claim 2 : The excenter of $\triangle{DEF}$ is the orthocenter of obtuse $\triangle{ABC}$.
Proof for claim 2 :
Let $H$ be the orthocenter of $\triangle{ABC}$. We may suppose that $\angle{ACB}$ is obtuse. Since $\triangle{ABD}$ is a right triangle with $\angle{ADB}=90^\circ$, $\angle{BAD}$ is acute. Similarly, $\angle{ABE}$ and $\angle{BHD}$ are acute. It follows from these that $\triangle{ABH}$ is an acute triangle. $FC, EC, DC$ is an angle bisector of $\angle{EFD},\angle{FED},\angle{FDE}$ respectively. Since $EC\perp EH$ and $DC\perp DH$, we see that $EH, DH$ is an exterior angle bisector of $\angle{FED},\angle{FDE}$ respectively, so the claim follows.$\quad\square$
Claim 3 : If $D,E,F$ are three distinct points, then the number of $\triangle{ABC}$ is $4$.
Proof for claim 3 :
$\triangle{ABC}$ is a right triangle if and only if either $D=E, E=F$ or $F=D$ holds. So, if $D,E,F$ are three distinct points, then we see that $\triangle{ABC}$ is either an acute triangle or an obtuse triangle. It follows from claim 1 that there is only one acute $\triangle{ABC}$. Also, it follows from claim 2 that there are only three obtuse $\triangle{ABC}$. Therefore, the claim follows.$\quad\square$