Analytic Geometry Questions

Question

1. Find the largest real number $x$ for which there exists a real number $y$ such that $x^2 + y^2 = 2x + 2y$.

2. In triangle $ABC$, $AB = AC$, $D$ is the midpoint of $\overline{BC}$, $E$ is the foot of the perpendicular from $D$ to $\overline{AC}$, and $F$ is the midpoint of $\overline{DE}$. Prove that $\overline{AF}$ is perpendicular to $\overline{BE}$.

For #1, i have put the equation into the circle equation form, but know not what to do next.

Answer 1

2 problem: ask to prove AF Perpendicular to BE.

Solution:

I added three lines BG perp to AC, H is the midpoint of CE. FK is an extension of AF.

(1) BG is parallel to DE; (because both perp to AC).

(2) E is the midpoint of CG; (because D is the midpoint of BC).

(3) BE is parallel to DH; ( because H is the midpoint of CE; and look at similarity of the triangle BCG and DCE).

(4) Now the question turns to: if DH is perp to AK, the problem will be solved! now let's focus on this.

(5) Triangle ADC is similar to DCE; and ADC is also similar to ADE. So

(6) Triangle DCE is similar to ADE!

(7) H is the midpoint at edge CE, and F is the midpoint at edge DE. So

(8) Triangle AFE is similar to DEH! so

(9) Angle EDH = angle FAE.

(10) Since angle EDA + DAE = 90 deg; and

(11) Angle DAE = angle DAF + FAE = angl DAF + angle EDH = angle HDA!

(12) Angle HDA + DAF(or DAK) = 90 deg; (from 10 and 11)

(13) so angle DKA = 90 deg. (because the sum of the angles of a triangle is 180 deg).

So AK is perp to DH.

Above is a brief idea for the prove.

Answer 2

1: The equation can be written

(x-1)^2 + (y-1)^2 = 2;

So (y-1)^2 >= 0

Take its minimum,

(x-1)^2 = 2;

Solve it

 x = 1 + sqrt(2); 

This is the maximum x can take while y is a real number.

Answer 3

Look at $x^2 + y^2 = 2x + 2y$ as a quadratic equation in $y$,

$$y^2-2y+x^2-2x=0.$$

This equation has real solutions when the discriminant is positive, i.e.

$$4-4(x^2-2x)\ge0.$$

This equation in $x$ has the roots $1\pm\sqrt2$ and the expression is positive between the roots. Then the largest $x$ is the largest root.