# Math Help - Hard math problem

1. ## Hard math problem

My friend gave me a good question today so I want to see if you smart geniuses can answer it

Consider a fixed line AB=4. Also consider a line CD, such that
AB=CD=4, and that AB is the perpendicular bisector of CD AND CD is the perpendicular bisector of AB.
Now, consider the set of all points P on CD (which there are inifineitely many). For every unique point P, there is a unique point Q such that QBPA is cyclic (all four points lie on a circle).
The set of all possible points Q creates a familiar figure.
What is that figure and what is the area of that figure?
Prove this.

I said it was a cirlce with area 4pi, but he said it was wrong?
What other familar shape can it possibly be? If i draw it it seems like a circle (i just plotted a lot of points)

2. Hello,

For every unique point P, there is a unique point Q such that QBPA is cyclic
Is there some information missing here ?
Because for a point P, there are infinitely many points Q such that Q,B,P,A lie on a circle

Or am I missing something ?

3. Sorry, same user, i forogt user name

anyways, you are right

I forgot one more condition (forogt what he tole me):

QP is bisected by AB. Thus there are two unique points Q for every P.

Anyways, inifintiely many points P will create a unique figure for P.

What is the figure and what is that area

4. Originally Posted by mluo
Consider a fixed line segment AB=4. Also consider a line segment CD, such that
AB=CD=4, and that AB is the perpendicular bisector of CD AND CD is the perpendicular bisector of AB.
Now, consider the set of all points P on CD (which there are inifineitely many). For every unique point P, there are two unique points Q such that QBPA is cyclic (all four points lie on a circle) and that QP is bisected by AB.
The set of all possible points Q creates a familiar figure.
What is that figure and what is the area of that figure?
Prove this.

I said it was a cirlce with area 4pi, but he said it was wrong?
What other familar shape can it possibly be? If i draw it it seems like a circle (i just plotted a lot of points)
1. Please don't double post. It's against the rules and it wastes everybodies time.

2. See attachment

5. It is an ellipse, with foci at A and B and eccentricity 1/√2. So its area is 4√2π.

Now prove it.

6. Isnt the proof just Angle Bisector or am I wrong?
This is because the chords AP and BP equal.
Do you have a good solution?

7. Originally Posted by person8901
Isnt the proof just Angle Bisector or am I wrong?
This is because the chords AP and BP equal.
Do you have a good solution?
I have a solution, though I wouldn't call it a "good" solution. (I would prefer to have an argument using synthetic geometry to show that AQ+QB = const.)

Take a coordinate system in which A and B are the points (±a,0), and C and D are (0,±a). (We are told that a=2, but I prefer to work with letters rather than numbers.)

If a circle contains A, B and P, then its centre must be on the y-axis, say at the point (0,t). Then the radius of the circle is $\sqrt{t^2+a^2}$, and its equation is $x^2 + (y-t)^2 = t^2+a^2$. Thus P is the point $(0,t-\sqrt{t^2+a^2})$.

The condition that PQ is bisected by AB tells you that the y-coordinate of Q is the negative of the y-coordinate of P. So if Q=(x,y) then $y = \sqrt{t^2+a^2} - t$. Putting those coordinates into the equation of the circle, and simplifying, we get the equation $x^2 = 4t(\sqrt{t^2+a^2}-t) = 4ty$. Thus $t = \frac{x^2}{4y}$. Substitute that value of t into the equation $y = \sqrt{t^2+a^2} - t$, and you get $y = \sqrt{\frac{x^4}{16y^2} + a^2} - \frac{x^2}{4y}$. Write this as $4y^2+x^2 = \sqrt{x^4 + 16a^2y^2}$, square both sides, simplify, and the result comes out as $\frac{x^2}{2a^2} + \frac{y^2}{a^2} = 1$, the standard equation of an ellipse with semi-axes √2a and a.