Locate The A B

Geometry Level 5

The following figure shows a cyclic quadrilateral $ABCD$ in the $xy$ -plane, with line segment $AC$ whose equation is $x-y-1=0$ and diameter $BD$ . Points $M(0;4)$ and $N(2;2)$ are the foots of the perpendicular lines from point $A$ to line segments $BC$ and $BD$ .

If the coordinates of points $A$ and $B$ can be represented as $(x_a;y_a)$ and $(x_b;y_b)$ respectively and $x_a<2$ , find $x_a+y_a+x_b+y_b$ .

Note: The above figure is not drawn to scale

The answer is 2.

1 solution

Leah Jurgens
Jun 11, 2017

See image above for details

Line segment $MN$ has an equation of $x+y-4=0$ .

Now let line segment $MN$ cuts $AC$ at point $P$ . We can imply the coordinates of $P$ that is $(\frac{5}{2};\frac{3}{2})$ .

We have $AM \parallel DC$ and $ABMN$ is a cyclic quadrilateral so:

$\angle PAM = \angle PCD = \angle ABD = \angle AMP$

$\Rightarrow PA=PM$

Because $A \in AC: x-y-1=0$ so let $A(x_a;x_a-1)$

We have: $PA = PM \Rightarrow x_a=0 or x_a=5$ . $x_a=0<2$ so $A(0;-1)$ .

We can easily find the equation of line segments $BD$ and $BC$ because they pass through point $N$ and $M$ and are perpendicular to $AN$ and $AM$ . Therefore, we can find the coordinates of point $B$ that is $(-1;4)$ .

In conclusion, $x_a+y_a+x_b+y_b=2$

Locate The A B

The answer is 2.

1 solution

0 pending reports