## Thursday, May 21, 2015

### From a friend

From a friend

If $f(x) = \sec(x^{333})$ then find $f^{666}(x)$ Where $f^n(x)$ represents the $n$-th derivative of a function.

Solution

At first this looks crazy but we quickly notice that a lot of terms in the 666-th derivative will evaluate to zero at $x=0$. More precisely, the first derivative contains factors $\sec(x^{333}), \tan(x^{333}),$ and $x^{332}$.

The tangent and the $x^n$ factors evaluate to zero, but the secant factor evaluates to 1. So all we need to figure out is, at each derivative step, which terms will affect the final derivatives.

To make it clearer, let's generalize into: $f(x) = \sec(x^n)$. $$f^1(x) = nx^{n-1} \sec(x^n) \tan(x^n)$$ $$f^2((x) = n^2x^{2n-2} \sec(x^n) \tan^2(x^n) + n(n-1)x^{n-2} \sec(x^n) \tan(x^n) + n^2 x^{2n-2} \sec^3(x^n)$$ The first term will continue to have higher derivatives with higher powers of $x$ and $\tan(x^n)$. Ditto for the second term. What we need to worry about is only the third term.

## Tuesday, May 5, 2015

### Tangents and secant

On a circle $L_1$ centered at $O$, we draw two points $A$ and $B$ that are not diametrically opposite. The tangents at $A$ and $B$ meet at $P$. Circle $L_2$ are created with $OB$ as the diameter. $L_2$ and $AB$ intersect at $B$ and $Q$. Show that $Q$ lies on $OP$.

### four tangent circles

Given four circles: $L_1,L_2,L_3,L_4$ such that $L_1$ is tangent to $L_2$ at $X$, $L_2$ is tangent to $L_3$ at $Y$, $L_3$ is tangent to $L_4$ at $Z$ and $L_4$ is tangent to $L_1$ at $W$. All tangencies are outside (the circles do not overlap each other other than at the points of tangency).

Show that $X,Y,Z,W$ all lie on a circle.

## Monday, May 4, 2015

### Excircle and circumcircle

In $\triangle ABC$, the $L_1$ be the excircle with respect to $AB$. It touches $CA$ and $CB$ and $M$ and $N$. The circle centered at $C$ and passing through $M$ and $N$ intersects $AB$'s extension at $P,Q$. Show that $L_2$ the circumcircle of $\triangle CPQ$ is tangent to $L_1$, and that the point of tangency between $L_1$ and $L_2$, the point of tangency between $L_1$ and $\triangle ABC$, and $C$ all lie on one line.

Solution Let $X$ be the tangency point of $L_1$ and $AB$. Let $CX$ meets $L_1$ in $Y_1$ and $CX$ meets $L_2$ in $Y_2$. We shall establish that $Y_1 = Y_2$.

Let $C'$ be the diameter of $L_2$ and let $CC'$ meets $AB$ (or its extension) at $D$. Because $CP = CQ$ then it's easy to see that $CD \perp PQ$. Let $s = CM = CN = CP = CQ$.

By power point on $L_1$: $$CX.CY_1 = CM^2 = s^2$$.

By similarity of $\triangle CXD \sim \triangle CC'Y_2$ and $\triangle CC'P \sim \triangle CDP$ : $$CX.CY_2 = CD.CC' = CP^2 = s^2$$

Thus $CY_1 = CY_2$ which means $Y_1 = Y_2$. Now we know that $L_1$ meets $L_2$ in at least one point $Y = Y_1 = Y_2$. Suppose they meet in another point $Z \neq Y$

Let $CZ$ meets $L_1$ in $CZ$ and $X_1 \neq X$. By power point we again have:$CX_1.CZ = s^2$.

Let $CZ$ meets $AB$ in $X_2$. Similar as above, because $Z \in L_2$, we can repeat the similarity argument to establish: $CX_2.CZ = s^2$. This means that $CX_1 = CX_2$, thus $X_1 = X_2 \in L_1, AB$ but that is a contradiction because $L_1$ is only tangent to $AB$, and can only meet $AB$ at $X$.

Now that we've established the tangency between $L_1$ and $L_2$, it's clear that $C,X,Y_1/Y_2$ all lie on one line from the proof.

### Circumcenter and 3 circumcircles

Given an acute triangle $ABC$ and its circumcircle is centered at $O$ and has radius $R$.

$AO$ meets the circumcircle of $\triangle OBC$ at $X$, $BO$ meets the circumcircle of $\triangle OAC$ at $Y$, and $CO$ meets the circumcircle of $\triangle OAB$ at $Z$.

Show that $$AX + BY + CZ \geq 9R$$

Determine when equality occurs.

Solution

Let $AX$ meets $BC$ at $D$. And let $OO'$ be the diameter of the circumcircle of $\triangle OBC$, and let $M$ be the intersection of $OO'$ and $BC$.

$$\frac{OX}{OO'} = \frac{OM}{OD}$$ $$OX.OD = OO'.OM = OC^2 = R^2$$ $$AX = AO + OX = R + \frac{R^2}{OD} = \frac{R(R + OD)}{OD} = R\frac{AD}{OD}$$

If we define $E$ as the intersection of $BY$ and $AC$, and $F$ as the intersection of $CZ$ and $AB$, then similarly: $$BY = R \frac{BE}{OE}$$ $$CZ = R \frac{CF}{OF}$$

So: $$AX + BY + CZ = R(\frac{AD}{OD} + \frac{BE}{OE} + \frac{CF}{OF})$$ $$\geq \frac{9R}{\frac{OD}{AD} + \frac{OE}{BE} + \frac{OF}{CF}}$$ But $OD / AD = A_{\triangle OBC} / A_{\triangle ABC}$ and so on, so: $$\frac{OD}{AD} + \frac{OE}{BE} + \frac{OF}{CF} = 1$$ $$AX + BY + CZ \geq 9R$$

and equality happens when the area of $\triangle OBC$ is exactly $\triangle ABC / 3$. And this happens only when $O$ is the centroid of $\triangle ABC$, which means $ABC$ is an equilateral.

## Wednesday, April 8, 2015

### chessboard manhattan move

Given a $2015 \times 2015$ chessboard, a pawn is placed at the bottom left corner. At each turn, the pawn may move one, two, or three squares to the right or to the top, and the goal is to reach the top right corner. The rules are: If the previous move was to the top, then the next move must be to the right. If the previous move was to the right, then the next move must be to the top. The pawn may not move to the left or to the bottom at any point. The difference of length between each two adjacent moves must be exactly 1. How many legal paths are there for the pawn to reach the top right corner?

### Pawns on infinite chessboard

Let $k$ be a fixed integer. The cells in a $1 \times \infty$ chessboard is colored in the following fashion: ..., red, blue, green, red, blue, green, ... (and so on) There are a number of pawns on the board. At each turn, we are allowed to choose two pawns $A,B$ and make $A$ "jump over" $B$ so that the new distance is $k$ times the old distance. Formally, we may choose $A$ and $B$ with distance $d$ and move $A$ so that the new distance is $kd$ and $A$ is on a different side of $B$. Multiple pawns may occupy a single cell. Determine all values of $k$ such that, regardless of initial configurations, it's always possible to move the pawns to all occupy the same color.

### Circles on a sphere

Given a sphere of radius $R$, we draw 2015 circles with radius $R$ on it, such that there are no three circles that all meet in one point. It's easy to see that there will be $2^{2015}$ regions created on the sphere surface. Prove that we can color each region with black or white so that no two adjacent regions are colored the same and the total area painted black is the same as the total area painted white.