A hypotrochoid is a roulette traced by a point attached to acircle of radius r rolling around the inside of a fixed circle of radius $R$, where the point is a distance from the center of the interior circle.The parametric equations for a hypotrochoid are:
\[x=(R-r)\cos\theta+d\cos\left(\frac{R-r}{r}\theta\right)\]
\[y=(R-r)\sin\theta-d\sin\left(\frac{R-r}{r}\theta\right)\]

Where $\theta$ is the angle formed by the horizontal and the center of the rolling circle (note that these are not polar equations because $\theta$ is not the polar angle).

Special cases include the hypocycloid with $d=r$ and theellipse with $R=2r$

The classic Spirograph toy traces out hypotrochoid andepitrochoid curves
.

APMO Geometry

Let $AD$ be the distance from $A$ to $BC$, $H$ is the orthocenter of $\Delta ABC$
,$M$ is the middle point of $BC$,MH∩ABC=E ED∩○ABC=F

show that $AB:AC=BF:CF$

Here is a picture:

Let I be the intersection of EM and the circle，It is not hard to prove $HBIC$ is a parallelogram. By the law of sine, we have：

\[\angle{BAD}=\alpha\,\,\angle{DAC}=\beta%20\\\frac{BD}{\sin{\alpha}}=AB\,\,\,\,\frac{CD}{\sin{\beta}}=AC\]

\[\frac{\sin\beta}{HM}=\frac{\sin\angle{BHM}}{BM}\,\,\,\,\,\,\frac{\sin\alpha}{HM}=\frac{\sin\angle{HMC}}{MC}\iff\frac{\sin\beta}{\sin\alpha}=\frac{\sin{BHM}}{\sin{MHC}}\]

\[\frac{BF}{CF}=\frac{BD}{CD}\cdot\frac{\sin\beta}{\sin\alpha}=\frac{AB}{AC}\]

Problem with rectangle

 If a finite number of rectangles, every one of which has at least one integer side, perfectly tile a big rectangle, then the big rectangle also has at least one integer side.

An interesting proof

roof by complex integration

A standard proof of theorem 1 (IBM 1999b) involves integrating the complex function e^2πi (x+y) over the two-dimensional plane. The integral over any rectangle is zero if and only if at least one side is an integer; so the integral over each small rectangle is zero; the integral over the big rectangle is the sum of the small rectangles' integrals, so it equals zero too; therefore the big rectangle has an integer side.

This proof is probably not accessible to most ten-year olds; nor is the second solution given at (IBM 1999b), which involves large prime numbers. And neither of these proofs helps prove theorem 3.

I now present two proofs of theorem 1, both accessible to ten-year-olds. The first proof is similar in character to the complex integral proof sketched above. The second proof uses a completely different approach, which can be applied immediately to theorem 3. Both proofs have been published before (Wagon, 1987).

Figure 1. Illustration of the checkerboard proof. Dashed blue lines are an integer grid. The checkerboard squares are of size 1/2. In the lefthand example (a), the central small rectangle violates the constraint of the theorem: it has both sides non-integer. In the righthand example (b), every small rectangle has one side of integer length. Every small rectangle covers equal amounts of black and white, so the large rectangle must do the same.

Geometry Problem from China high school mathematics tournament

Let $AB$ be a chord in the circle $A_{1}$; $AE=EF=FB$; Point $X$ is on the circle, $PE$ meet with the circle at point $Y$, $PF$ meet with the circle at point $Z$. Show that $AY\cdot BZ=EF\cdot YZ$.

Proof:  Apply the law of sine, we have

\[\frac{\sin{\angle AZY}}{\sin{\angle{YAZ}}}=\frac{AY}{YZ}=\frac{\sin{\angle AXY}}{\sin{\angle YXZ}}\]

\[=\frac{XF}{XA}=\frac{\sin\angle XAE}{\sin\angle XFA}=\frac{BF}{BZ}=\frac{EF}{BZ}\]

Comment: This problem is not kind of hard. It can be proved with pure geometry method easily too. Just consider the rotation of $\Delta AXZ$  

prove the minimum value by completing square

let $x,y$be positive poof that \[x^3+y^3+xy^2+x^2y-2x^2-2y^2-2xy+2\geqslant0\]

Poof: it can be written as \[\frac{1}{4}(x-1)^2(3x+y)+\frac{1}{4}(y-1)^2(3y+x)+\frac{1}{4}(x+y+2)(x-2+y)^2\], done

The product of 3 consecutive positive integers is never a square.

Proof: if we have $a(a^2-1)=b^2$, then$\gcd(a,a^2-1)=\gcd(a^2-1,a^2)=1$, so$a=m^2, a^2-1=n^2$, where$mn=b$, and$\gcd(m,n)=1$. But $a^2-1=n^2$ has no integer solution.

\sqrt[n]{n}

Let $n$ be an positive integer, prove that \[\sqrt{n}\leqslant\sqrt[n]{n!}\leqslant\frac{n+1}{2}\]