number of such triangles, if feet of perpendicular are given

Question

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

Answer 1

The answer is $4$.

We can say that if $D,E,F$ are three distinct points, then the answer is $4$.

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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 $4$.

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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.$\quad\square$

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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.$\quad\square$

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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.$\quad\square$

Answer 2

There will be, first of all (see EDIT at bottom, thanks to Blue for his useful comment), a triangle $\triangle ABC$, whose orthocenter $H$ is the incenter of the orthic triangle $\triangle DEF$. We can find the coordinates of $A$, $B$, and $C$, as follows.

1. Equation of line $DE$: $$r_{DE} \ : \ 3x-4y+40 = 0$$
2. Equation of line $EF$: $$r_{EF} \ : \ x - 8.$$
3. Equation of line $DF$: $$r_{DF} \ : \ 4 x-3y -5=0 $$
4. Equation of altitude $AD$. This is one of the bisectors of the angles formed by $r_{DE}$ and $r_{DF}$. We can find these using the formula $$\frac{|3x-4y+40|}{\sqrt{9+16}}=\frac{|4x-3y-5|}{\sqrt{9+16}}.$$The correct solution, in this case, is the one with positive slope, i.e.$$r_{AD} : x-y+5=0.$$
5. Side $BC$ will be perpendicular to $AD$, yielding $$r_{BC}\ : \ x+y-45=0.$$
6. Analogously, $BE$ is one of the bisectors of the angles formed by $r_{EF}$ and $r_{DE}$, that is $$\frac{|x-8|}{1}=\frac{|3x-4y+40|}{5}.$$The solution with negative slope is $$r_{BE} \ : \ x+2y-40=0.$$
7. Using 5. and 6. we find $B= r_{BE}\cap r_{BC}$, that is $\boxed{B(50,-5)}$.
8. Side $r_{AB}$ is the line through $B$ and $F$, that is $$r_{AB} \ : \ x+3y-35=0.$$
9. Use 4. and 8. to find $A=r_{AD} \cap r_{AB}$, i.e. $\boxed{A(5,10)}$.
10. Side $AC$ is the line through $A$ and $E$, meaning $$r_{AC} \ : \ 2x-y=0.$$
11. Finally we get $C= r_{AC} \cap r_{BC}$, using 5. and 10., which yield $\boxed{C(15,30)}$.

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EDIT

Other three triangles that share the same altitude feet: $\triangle HAB$, $\triangle HBC$, and $\triangle HAC$, where $H=r_{AD}\cap r_{BE}$ is the orthocenter of $\triangle ABC$.

Answer 3

The number is $4$.

We may use the fact that the dot product of two non-zero vectors is zero if and only if the two vectors are perpendicular.

Let $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$.

Since $D, E, F$ are three distinct points, we know that $A, B, C, D, E, F$ are six distinct points.

$A, B, C, D, E, F$ must satisfy the following six equations: \begin{align} (x_1 - 20)(x_2 - 20) + (y_1 - 25)(y_2 - 25) &= 0, \tag{1}\\ (x_1 - 20)(x_3 - 20) + (y_1 - 25)(y_3 - 25) &= 0, \tag{2}\\ (x_2 - 8)(x_1 - 8) + (y_2 - 16)(y_1 - 16) &= 0, \tag{3}\\ (x_2 - 8)(x_3 - 8) + (y_2 - 16)(y_3 - 16) &= 0, \tag{4}\\ (x_3 - 8)(x_1 - 8) + (y_3 - 9)(y_1 - 9) &= 0, \tag{5}\\ (x_3 - 8)(x_2 - 8) + (y_3 - 9)(y_2 - 9) &= 0. \tag{6} \end{align} Explanation: Equations (1) through (6) describe $\overrightarrow{AD} \perp \overrightarrow{BD}$, $\overrightarrow{AD} \perp \overrightarrow{CD}$, $\overrightarrow{BE} \perp \overrightarrow{AE}$, $\overrightarrow{BE} \perp \overrightarrow{CE}$, $\overrightarrow{CF} \perp \overrightarrow{AF}$, and $\overrightarrow{CF} \perp \overrightarrow{BF}$, respectively.

All solutions $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ of the above system of equations are given by:

i) $(5,10), (50, -5), (15,30)$;

ii) $(50, -5), (5,10),(10,15)$;

iii) $(15,30),(10,15),(5,10)$;

iv) $(10,15),(15,30),(50,-5)$;

v) $(8,41),(8,16),(-4/3,9)$;

vi) $(8,34),(53/4,16),(8,9)$;

vii) $(20,25),(8,16),(8,9)$;

viii) $(20,25),(85/8,25/2),(10/3,25/2)$.

Since $A, B, C, D, E, F$ are six distinct points, the solutions v), vi), vii) and viii) do not meet the requirement. One may check that i), ii), iii) and iv) are all indeed the solutions.