Question

The coordinates of the feet of the perpendicular from the vertices of a triangle ABC on the opposite sides are D (20, 25), E (8, 16), F (8,9) Then the sum of x-coordinates of all vertices of the triangle is

Answer 1

70

Answer 2

Let the vertices of the triangle be A, B, C. Let the feet of the perpendiculars (altitudes) from A, B, C to the opposite sides BC, CA, AB be D, E, F respectively. We are given the coordinates of the feet of the altitudes:

D = (20, 25) E = (8, 16) F = (8, 9)

A key property in geometry states that the vertices of the original triangle ABC are the excenters of the pedal triangle DEF. Specifically:

• Vertex A is the excenter of $\triangle DEF$ opposite to vertex D.
• Vertex B is the excenter of $\triangle DEF$ opposite to vertex E.
• Vertex C is the excenter of $\triangle DEF$ opposite to vertex F.

Let the coordinates of the vertices of the pedal triangle be $D(x_D, y_D)$, $E(x_E, y_E)$, $F(x_F, y_F)$. Let the side lengths of the pedal triangle be $d = EF$, $e = DF$, $f = DE$.

First, calculate the side lengths of the pedal triangle DEF using the distance formula: $d = EF = \sqrt{(x_E - x_F)^2 + (y_E - y_F)^2} = \sqrt{(8 - 8)^2 + (16 - 9)^2} = \sqrt{0^2 + 7^2} = 7$ $e = DF = \sqrt{(x_D - x_F)^2 + (y_D - y_F)^2} = \sqrt{(20 - 8)^2 + (25 - 9)^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20$ $f = DE = \sqrt{(x_D - x_E)^2 + (y_D - y_E)^2} = \sqrt{(20 - 8)^2 + (25 - 16)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15$

The coordinates of the excenter opposite to a vertex (say D) are given by the formula: $A = (x_A, y_A) = \left( \frac{-d x_D + e x_E + f x_F}{-d + e + f}, \frac{-d y_D + e y_E + f y_F}{-d + e + f} \right)$ $B = (x_B, y_B) = \left( \frac{d x_D - e x_E + f x_F}{d - e + f}, \frac{d y_D - e y_E + f y_F}{d - e + f} \right)$ $C = (x_C, y_C) = \left( \frac{d x_D + e x_E - f x_F}{d + e - f}, \frac{d y_D + e y_E - f y_F}{d + e - f} \right)$

We need to find the sum of the x-coordinates of all vertices of the triangle ABC, i.e., $x_A + x_B + x_C$.

Calculate the denominators for the x-coordinates: For $x_A$: $-d + e + f = -7 + 20 + 15 = 28$ For $x_B$: $d - e + f = 7 - 20 + 15 = 2$ For $x_C$: $d + e - f = 7 + 20 - 15 = 12$

Now, calculate the x-coordinates of A, B, and C: $x_A = \frac{-7(20) + 20(8) + 15(8)}{28} = \frac{-140 + 160 + 120}{28} = \frac{140}{28} = 5$ $x_B = \frac{7(20) - 20(8) + 15(8)}{2} = \frac{140 - 160 + 120}{2} = \frac{100}{2} = 50$ $x_C = \frac{7(20) + 20(8) - 15(8)}{12} = \frac{140 + 160 - 120}{12} = \frac{180}{12} = 15$

Finally, sum the x-coordinates: Sum of x-coordinates = $x_A + x_B + x_C = 5 + 50 + 15 = 70$.