Question

A triangle is so placed that the midpoints of its sides one placed on the co-ordinate axes, if a, b and c are the sides of the triangle, then the equation of the triangle is $\frac{x}{x_1}+\frac{y}{y_1}+\frac{z}{z_1}=1$ where $a^2+b^2+c^2=k(x_i^2+y_i^2+z_i^2)$, then the value of $k$ is ____.

Answer 1

8

Answer 2

Let the vertices of the triangle be $A(x_A, y_A, z_A)$, $B(x_B, y_B, z_B)$, and $C(x_C, y_C, z_C)$.
Let the midpoints of the sides $BC$, $CA$, and $AB$ be $D$, $E$, and $F$ respectively.
According to the problem statement, these midpoints are placed on the coordinate axes.
Let $D$ be on the x-axis, $E$ on the y-axis, and $F$ on the z-axis.
Let their coordinates be $D(x_1, 0, 0)$, $E(0, y_1, 0)$, and $F(0, 0, z_1)$.
(Note: The problem uses $x_1, y_1, z_1$ in the equation of the plane, which are consistent with these midpoint coordinates.)

Using the midpoint formula:
For midpoint $D$ of $BC$:
$D = \left(\frac{x_B+x_C}{2}, \frac{y_B+y_C}{2}, \frac{z_B+z_C}{2}\right) = (x_1, 0, 0)$
This gives us:

1. $x_B+x_C = 2x_1$
2. $y_B+y_C = 0$
3. $z_B+z_C = 0$

For midpoint $E$ of $CA$:
$E = \left(\frac{x_C+x_A}{2}, \frac{y_C+y_A}{2}, \frac{z_C+z_A}{2}\right) = (0, y_1, 0)$
This gives us:
4) $x_C+x_A = 0$
5) $y_C+y_A = 2y_1$
6) $z_C+z_A = 0$

For midpoint $F$ of $AB$:
$F = \left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}, \frac{z_A+z_B}{2}\right) = (0, 0, z_1)$
This gives us:
7) $x_A+x_B = 0$
8) $y_A+y_B = 0$
9) $z_A+z_B = 2z_1$

Now, we solve for the coordinates of the vertices $A, B, C$:

From (4), $x_C = -x_A$.
From (7), $x_B = -x_A$.
Substitute $x_B = -x_A$ and $x_C = -x_A$ into (1):
$(-x_A) + (-x_A) = 2x_1 \Rightarrow -2x_A = 2x_1 \Rightarrow x_A = -x_1$.
Then, $x_B = -(-x_1) = x_1$ and $x_C = -(-x_1) = x_1$.
So, $x_A = -x_1$, $x_B = x_1$, $x_C = x_1$.

From (2), $y_C = -y_B$.
From (8), $y_A = -y_B$.
Substitute $y_C = -y_B$ and $y_A = -y_B$ into (5):
$(-y_B) + (-y_B) = 2y_1 \Rightarrow -2y_B = 2y_1 \Rightarrow y_B = -y_1$.
Then, $y_A = -(-y_1) = y_1$ and $y_C = -(-y_1) = y_1$.
So, $y_A = y_1$, $y_B = -y_1$, $y_C = y_1$.

From (3), $z_C = -z_B$.
From (6), $z_A = -z_C$.
Substitute $z_C = -z_B$ and $z_A = -z_C = -(-z_B) = z_B$ into (9):
$z_B + z_B = 2z_1 \Rightarrow 2z_B = 2z_1 \Rightarrow z_B = z_1$.
Then, $z_A = z_1$ and $z_C = -z_1$.
So, $z_A = z_1$, $z_B = z_1$, $z_C = -z_1$.

Thus, the coordinates of the vertices are:
$A(-x_1, y_1, z_1)$
$B(x_1, -y_1, z_1)$
$C(x_1, y_1, -z_1)$

Now, we calculate the squares of the side lengths $a, b, c$:
$a^2 = BC^2 = (x_1-x_1)^2 + (-y_1-y_1)^2 + (z_1-(-z_1))^2$
$a^2 = 0^2 + (-2y_1)^2 + (2z_1)^2 = 4y_1^2 + 4z_1^2$

$b^2 = CA^2 = (x_1-(-x_1))^2 + (y_1-y_1)^2 + (-z_1-z_1)^2$
$b^2 = (2x_1)^2 + 0^2 + (-2z_1)^2 = 4x_1^2 + 4z_1^2$

$c^2 = AB^2 = (-x_1-x_1)^2 + (y_1-(-y_1))^2 + (z_1-z_1)^2$
$c^2 = (-2x_1)^2 + (2y_1)^2 + 0^2 = 4x_1^2 + 4y_1^2$

Now, sum $a^2+b^2+c^2$:
$a^2+b^2+c^2 = (4y_1^2 + 4z_1^2) + (4x_1^2 + 4z_1^2) + (4x_1^2 + 4y_1^2)$
$a^2+b^2+c^2 = (4x_1^2+4x_1^2) + (4y_1^2+4y_1^2) + (4z_1^2+4z_1^2)$
$a^2+b^2+c^2 = 8x_1^2 + 8y_1^2 + 8z_1^2$
$a^2+b^2+c^2 = 8(x_1^2 + y_1^2 + z_1^2)$

The problem states that $a^2+b^2+c^2=k(x_i^2+y_i^2+z_i^2)$.
Comparing our result with the given equation, we find that $k=8$.
The equation of the triangle $\frac{x}{x_1}+\frac{y}{y_1}+\frac{z}{z_1}=1$ represents the plane containing the triangle, and the points $D(x_1,0,0)$, $E(0,y_1,0)$, $F(0,0,z_1)$ are indeed the intercepts of this plane with the coordinate axes. Also, the vertices $A, B, C$ satisfy this plane equation.