Triangle Inequality

In a triangle $ABC$, let $a = BC, b = AC, c = AB$. For any point $M$ and real numbers $x,y,z$, show that

$(x+y+z)(xMA^2 + yMB^2 + zMC^2) \geq xyc^2 + xzb^2 + yza^2$

Solution
We shall show that the inequality above is equivalent to
$$(x \vec{MA} + y \vec{MB} + z \vec{MC})^2 \geq 0$$

Indeed, since:
$$2\vec{MA} \vec{MB} = MA^2 + MB^2 - c^2$$
so
$$2xy\vec{MA} \vec{MB} = xyMA^2 + xyMB^2 - xyc^2$$
$$2yz\vec{MB} \vec{MC} = yzMB^2 + yzMC^2 - yza^2$$
$$2zx\vec{MC} \vec{MA} = zxMC^2 + zxMA^2 - zxb^2$$

Adding them, we obtain:
$$RHS = \sum x(y+z)MA^2 - 2\sum xy \vec{MA} \vec{MB}$$

So
$$LHS - RHS = \sum x^2MA^2 + \sum x(y+z)MA^2 - RHS = \sum x^2 MA^2 + 2\sum xy \vec{MA} \vec{MB}$$
$$= (x \vec{MA} + y \vec{MB} + z \vec{MC})^2 \geq 0 $$

Combined Sequence

Let $A,B$ be two distinct positive integers greater than 1, and define the sequences:

$a_m = m + \frac{Bm}{A}, m = 1,2,3,\cdots, A-1$
$b_m = m + \frac{Am}{B}, m = 1,2,3,\cdots, B-1$

And combine those two sequence to form a new sequence $c_1,c_2,\cdots, c_{A+B-2}$ with $c_1 \leq c_2 \leq \cdots \leq c_{A+B-2}$.

Prove that the difference between any two consecutive $c_i$s are less than 2.

Solution

It suffices to prove that for each $a_k$, we can find another $a_i$ or $b_j$ that lies between $a_k$ and $a_k+2$, and vice versa, for each $b_l$, we can find another $a_i$ or $b_j$ that lies between $b_l$ and $b_l + 2$.
Without loss of generality, we may assume that $A > B$.
The distance between two consecutive $b_j$s are $b_{j+1} - b_j = 1 + B/A < 2$, so for each $b_l$, we're guaranteed that $b_l < b_{l+1} < b_l+2$.

Now we're left to consider $a_k$.
Fix $k$, and now take the smallest $l$ such that $b_l > a_k$. This is always possible because:
\[a_k \leq a_{B-1} = (B-1)(1+A/B) < (A-1)(1+B/A) = b_{A-1}\]
(the middle inequality is true because $A > B$.)

If $l$ is the smallest such $l$, that means
\[b_{l-1} \leq a_k \iff (l-1)(1+B/A) \leq k(1+A/B) \iff l/A - k/B \leq 1/A\]

So
\[b_l - a_k = l(1+B/A) - k(1+A/B) = (A+B)(l/A - k/B) \leq (A+B)/A < 2\]
which completes the proof

3 arbitrary functions

Suppose $f,g,h$ are functions that are defined in the closed interval $0 \leq x \leq 1$. Show that we can always find $a,b,c \in [0,1]$ such that:

$$|f(a)+g(b)+h(c) - (1-a)(1-b)(1-c)| \geq \frac{1}{3}$$

Also show that the constant $\frac{1}{3}$ cannot be replaced by a larger constant.

Finite Prime Set

Let $P$ be a finite set of primes. Define $m(P)$ as the largest number of consecutive integers each of which is divisible by a prime in $P$.
Let $|P|$ denote the size of $P$ and $\min(P)$ to be the smallest element in $P$.

Prove that:
1. $m(P) \geq |P|$
2. $m(P) = |P|$ if and only if $\min (P) > |P|$

Solution

Let $s = |P|$. We proceed by constructing $s$ consecutive integers which are each divisible by a prime in $P$. Indeed, the following system has a solution, according to Chinese Remainder Theorem:
$x+1 \equiv 0 \pmod {p_1}$
$x+2 \equiv 0 \pmod {p_2}$
...
$x+s \equiv 0 \pmod {p_s}$

Second part:
Suppose we have $s < \min(P)$ and that we have $s+1$ consecutive integers all divisible by some $p_i$: $x, x+1, x+2, ... , x+s$. Let $p_1 = \min(P)$.

For each $p_i$, it can only divide at most one of the above-mentioned integers. For if it divides $x+a$ and $x+b$ then it also divides $|a-b| \leq s < p_1$, a contradiction (since $p_1$ is the smallest prime in $P$).
So $m(P) \leq s$. But it's been shown that $m(P) \geq s$, so $m(P) = s$.

Now suppose we have $s \geq \min(P)$. We will show $s+1$ consecutive integers such that each is divisible by a prime in $P$, which then establishes $m(P) > s$.
Let $k = \min(P)$.
Again, we set up a system of equations as described above, but we choose the ordering of $p_i$s such that $p_k = k$. The rest can be arbitrary ordering. This is always possible if $s \geq k$.
According to CRT, there is a solution of $s$ consecutive integers that satisfy the above system of equations. But then we also have $x = x+k - k$ divisible by $p_k = k$. So together with $x$, they form $s+1$ consecutive integers.

Chess Board Coloring

Each cell in an infinite chess board is colored with one of the $n$ available colors. Prove that we can always find a rectangle such that all four corners have the same color.

Advanced version: Suppose not all cells are colored, but only some of them. Furthermore, for any circle with radius R, we can always find a colored cell in that circle. Prove that we can still find a monochromatic rectangle.

Solution: Cyclic quadrilateral

Original Problem: http://dharmath.blogspot.com/2010/05/cyclic-quadrilateral.html

First Solution

Let $AB=AD=x, BC=y$ and $CD=x+z$ with $y < z$.

Let $\theta = \angle ADC$ so $\angle ABC = 180^o-\theta$. It's easy to see that because $CD > BC$ then $\angle ABC > \angle ADC$ thus $0 < \theta < 90^o$

Now $AC^2 = x^2 + y^2 + 2xy \cos \theta = x^2 + (x+z)^2 - 2x(x+z)\cos \theta$, simplify it to get:
$\cos \theta = \frac{(x+z)^2-y^2}{2x(x+y+z)} = \frac{x+z-y}{x}$

But since $\theta$ is an acute angle, $\theta < 60^o \iff \cos \theta > 1/2$
So we need to show
$x+z-y > x$ which is true because $z > y$

Second Solution



Let $R$ be a point on $CD$ such that $CB = CR$ (see the picture).

Since $AB=AD$, then $CA$ is an internal angle bisector, and thus $\triangle ABC$ are congruent to $\triangle ARC$. That means $RD = CD - CR = CD - CB > AB = AD$.
Also, $AD = AB = AR$.
So $\triangle ARD$ is an isosceles where $AD = AR$ and $RD > AD$, which means that $\angle ADR < 60^o$ and thus $\angle ABC > 120^o$

Cyclic quadrilateral

In a cyclic quadrilateral $ABCD$ such that $AB = AD$ and $AB+BC < CD$, show that $ABC > 120^o$

Solution: http://dharmath.blogspot.com/2010/05/solution-cyclic-quadrilateral.html