Question

f combined equation of two sides AB and AC of a triangle is x2 – xy – 2y2 + x + y = 0 and (1, 1) is orthocenter, then circum-radius of ΔABC is

Answer 1

$\frac{5\sqrt{2}}{3}$

Answer 2

The combined equation of two sides AB and AC of a triangle is given by $x^2 - xy - 2y^2 + x + y = 0$.

First, we factorize this equation to find the individual equations of lines AB and AC.
The homogeneous part $x^2 - xy - 2y^2$ can be factored as $(x - 2y)(x + y)$.
So, the given equation can be written as $(x - 2y)(x + y) + (x + y) = 0$.
Taking $(x + y)$ common, we get $(x + y)(x - 2y + 1) = 0$.
Thus, the equations of the two lines are:
Line AB: $L_1 \equiv x - 2y + 1 = 0$
Line AC: $L_2 \equiv x + y = 0$

1. Find Vertex A:
Vertex A is the intersection point of lines AB and AC.
Solve the system of equations:
$x - 2y + 1 = 0 \quad (1)$
$x + y = 0 \quad (2)$
From (2), $y = -x$. Substitute this into (1):
$x - 2(-x) + 1 = 0$
$x + 2x + 1 = 0$
$3x + 1 = 0 \Rightarrow x = -1/3$
Then $y = -(-1/3) = 1/3$.
So, vertex A is $(-1/3, 1/3)$.

2. Use Orthocenter H to find other vertices:
The orthocenter H is given as $(1, 1)$.
The slope of line AB ($m_{AB}$) is $-(\text{coefficient of } x)/(\text{coefficient of } y) = -(1)/(-2) = 1/2$.
The slope of line AC ($m_{AC}$) is $-(1)/(1) = -1$.

The altitude from vertex C to side AB (let's call it $h_C$) is perpendicular to AB.
The slope of $h_C$ is $-1/m_{AB} = -1/(1/2) = -2$.
Since $h_C$ passes through the orthocenter H $(1, 1)$, its equation is:
$y - 1 = -2(x - 1)$
$y - 1 = -2x + 2$
$2x + y - 3 = 0 \quad (3)$

The altitude from vertex B to side AC (let's call it $h_B$) is perpendicular to AC.
The slope of $h_B$ is $-1/m_{AC} = -1/(-1) = 1$.
Since $h_B$ passes through the orthocenter H $(1, 1)$, its equation is:
$y - 1 = 1(x - 1)$
$y - 1 = x - 1$
$x - y = 0 \quad (4)$

3. Find Vertices B and C:
Vertex B is the intersection of line AB ($x - 2y + 1 = 0$) and altitude $h_C$ ($2x + y - 3 = 0$).
From $2x + y - 3 = 0$, we get $y = 3 - 2x$. Substitute this into $x - 2y + 1 = 0$:
$x - 2(3 - 2x) + 1 = 0$
$x - 6 + 4x + 1 = 0$
$5x - 5 = 0 \Rightarrow x = 1$
Then $y = 3 - 2(1) = 1$.
So, vertex B is $(1, 1)$.
Notice that vertex B $(1, 1)$ is the same as the orthocenter H $(1, 1)$.

4. Identify the type of triangle:
If a vertex of a triangle coincides with its orthocenter, then the triangle is a right-angled triangle, and the right angle is at that vertex.
In this case, B = H, so $\triangle ABC$ is right-angled at B.
We can verify this by checking the slopes of AB and BC.
Slope of AB ($m_{AB}$) = $1/2$.
Line BC is the altitude from C to AB, which we found as $2x + y - 3 = 0$. So, this is the equation of BC.
Slope of BC ($m_{BC}$) = $-2$.
Since $m_{AB} \cdot m_{BC} = (1/2) \cdot (-2) = -1$, AB is perpendicular to BC. Thus, $\angle B = 90^\circ$.

5. Find Vertex C:
Vertex C is the intersection of line AC ($x + y = 0$) and line BC ($2x + y - 3 = 0$).
From $x + y = 0$, we get $y = -x$. Substitute this into $2x + y - 3 = 0$:
$2x + (-x) - 3 = 0$
$x - 3 = 0 \Rightarrow x = 3$
Then $y = -3$.
So, vertex C is $(3, -3)$.

6. Calculate the Circum-radius R:
For a right-angled triangle, the circumcenter is the midpoint of the hypotenuse, and the circum-radius R is half the length of the hypotenuse.
The hypotenuse is AC.
Vertices are A $(-1/3, 1/3)$ and C $(3, -3)$.
Length of AC = $\sqrt{(3 - (-1/3))^2 + (-3 - 1/3)^2}$
$AC = \sqrt{(3 + 1/3)^2 + (-3 - 1/3)^2}$
$AC = \sqrt{(10/3)^2 + (-10/3)^2}$
$AC = \sqrt{100/9 + 100/9}$
$AC = \sqrt{200/9}$
$AC = \frac{\sqrt{200}}{3} = \frac{10\sqrt{2}}{3}$.

The circum-radius $R = AC/2$.
$R = \frac{10\sqrt{2}}{3} \div 2 = \frac{5\sqrt{2}}{3}$.

The final answer is $\boxed{\frac{5\sqrt{2}}{3}}$.