Osculating Circle, Ellipse, and Cone

An osculating circle of a point on a curve is defined as a circle that:
1. passes through that point
2. whose slope at that point is the same of the slope of the curve at that point
3. whose radius is the same as the radius of curvature of the curve at that point

In other words, it is a second-degree approximation circle of the curve at that point.

http://en.wikipedia.org/wiki/Osculating_circle

Given a cone whose half-angle is $\theta$, we take a cross section with a plane whose incident angle is $\theta$. That is, the plane is perpendicular to one of the cone rays. Naturally, the cross section forms an ellipse.

If O is the intersection of the main axis of the cone and the cross section, and A is the point on the ellipse's major axis that's closest to O, then prove that a circle with center O and radius OA is an osculating circle to the ellipse at A.

Hint: for people without any knowledge of calculus, the radius of osculating circle at A is $b^2/a$ where $b$ is half the length of minor axis and $a$ is half the length of major axis (standard ellipse notation).
The rest of the problem can be done without using calculus.

Solution:
http://dharmath.blogspot.com/2010/03/osculating-circle-ellipse-and-cone.html

Second Solution:
http://dharmath.blogspot.com/2010/03/second-solution-osculating-circle.html

Dissecting a tetrahedron

Let $T_r$ denote a regular tetrahedron whose side length is $r$.

1. Prove that it is impossible to assemble eight $T_1$ into a $T_2$.

2. Prove that a tetrahedron can be cut into four optically congruent pieces. Two solids $A$ and $B$ are considered optically congruent if either $A$ or its mirror image can be rotated into $B$.

3. Prove that a $T_2$ can be cut into four $T_1$ and a regular octahedron whose side length is 1.