Competitive Math Made Easy

 Let's take a look at some interesting sequences and series problems. 1. (USAMTS) Evaluate the value of  $$S = \sqrt{1+\cfrac{1}{1^2}+\cfrac{1}{2^2}} + \sqrt{1+\cfrac{1}{2^2}+\cfrac{1}{3^2}} + ... + \sqrt{1+\cfrac{1}{1999^2}+\cfrac{1}{2000^2}}$$ Solution Notice that $$1+\cfrac{1}{n^2}+\cfrac{1}{(n+1)^2} = \cfrac{n^4+2n^3+3n^2+2n+1}{n^2(n+1)^2} = \cfrac{n^2+n+1)^2}{n^2(n+1)^2} = \Bigg(1+\cfrac{1}{n(n+1)} \Bigg)^2$$ Thus, $$S= \Bigg(1+\cfrac{1}{1 \cdot 2} \Bigg) + \Bigg(1+\cfrac{1}{2 \cdot 3} \Bigg) + \Bigg(1+\cfrac{1}{3 \cdot 4} \Bigg) + ... + \Bigg(1+\cfrac{1}{1999 \cdot 2000} \Bigg)$$ $$= 1999 + \Bigg(1 - \cfrac{1}{2} \Bigg) + \Bigg(\cfrac{1}{2} - \cfrac{1}{3} \Bigg) + ... + \Bigg(\cfrac{1}{1999} - \cfrac{1}{2000} \Bigg) = 1999 + 1 - \cfrac{1}{2000} = \boxed{\cfrac{3,999,999}{2000}}$$ 2. (HMMT) Find the value of  $$S = \cfrac{1}{3^2+1} + \cfrac{1}{4^2+2} + \cfrac{1}{5^2+3} + ...$$ Solution For this problem, we can use partial fraction decomposition and try to find an alt...

Two triangles are called similar , if we can get two congruent triangles after enlarging or compressing the sides of one of them according to an equal ratio. That is, two triangles are similar means they have a same shape but may have different sizes. Criteria for Similarity of Two Triangles  (I) Each pair of corresponding angles are equal (A.A.A.);  (II) All corresponding sides are proportional (S.S.S.);  (III) Two pairs of corresponding sides are proportional, and the included corresponding angles are equal (S.A.S.);  (IV) For two right triangles, a pair of two corresponding acute angles are equal (A.A.);  (V) Among the three pairs of corresponding sides two pairs are proportional (S.S.). Basic Properties of Two Similar Triangles (I) For two similar triangles, their corresponding sides, corresponding heights, corresponding medians, corresponding angle bisectors, corresponding perimeter are all proportional with the same ratio.  (II) Consider the similarit...

When solving problems relating to clocks, keeping the following two points in mind will be helpful. The Minute hand covers $360^{\circ}$ in 60 minutes, or $6^{\circ}$ per minute. The Hour hand covers $360^{\circ}$ in 12 hours, or   $30^{\circ}$ per hour. Using these facts, the we can easily derive the results below: In a period of 12 hours, the hour and minute hand coincide 11 times. In a period of 12 hours, the hour and minute hand form a $180^{\circ}$ 11 times. In a period of 12 hours, any other angle between the minute and hour hand is formed 22 times. The time gap between any two coincidences is $\cfrac{12}{11}$ hours or $65\cfrac{5}{11}$ minutes. One common problem involving clocks is to find the angle between the minute and hour hand at a given time. To do so, one can use the formula  $$\theta = |5.5m - 30h |$$ Where $\theta =$ desired angle, and the time can be written as $h:m$. Let's use what we have learned to solve a couple problems. 1. What is the angle between the ...

Let $ABCD$ be a convex quadrilateral with $AB=BC=AC$. Suppose that a point P lies in the interior of the quadrilateral such that $AD=AP=DP$ and $\angle PCD=30^{\circ}$. Given that $CP=2$ and $CD=3$, find that length of $AC$.  The first step in solving this question is to draw the segment $PB$. Doing this allows us to see that $\triangle APB \cong \triangle ADC$ because $AP=AD$, $AB=AC$, and $\angle DAC = \angle PAB$ because $\angle DAP = \angle CAB = 60^{\circ}$. Thus, $PB=3$.  Next, let's set $\angle ACD = x$. We see that $\angle PCA = 30-x$, and $\angle BCP = 30 +x$. Due to the triangle congruency we established earlier, $\angle PBA = x$ as well. This means that we can write $\angle PBC = 60 - x$. Now,  $$\angle CPB = 180^{\circ} - \angle BCP - \angle PBC = 180^{\circ} - (30+x) - (60-x) = 90^{\circ}$$.  Thus, $\triangle BPC$ is a right triangle, and we know that $CP=2$ and $PB=3$, so we can easily find $CB$ to be  $$CB = \sqrt{2^2+3^2} = \boxed{\sqrt{13}}$$

The Pigeon Hole principle follows a simple observation that if you have 10 pigeons and 9 pigeonholes and you distribute these pigeons randomly, its quite obvious that there will be one pigeonhole with more than 1 pigeon. In other words, there are n boxes (the Pigeonholes), into which more than n objects (the Pigeons), have been placed, then at least one of the boxes must have received more than one of the objects. This principle is frequently applied in problems where we need to determine a minimal number of objects to ensure that some integral number property is satisfied. Consider the following puzzle:  A box contains red, black and white balls. The objective is to pick balls satisfying some constraints. How many balls must be taken to ensure that there is a pair of same color? The key to solve this puzzle is pigeonhole principle.  The principle says:  If $n$ pigeons are nesting in $m$ pigeonholes, where $n > m$, then at least one pigeonhole has more than one pig...

Given $\triangle ABC$ with incenter $I$, extend $AI$ to meet the circumcircle of $\triangle ABC$ at $L$. Then, reflect $I$ over $L$, and name this point $I_a$. Then:  $(i)$ $BICI_a$ is a cyclic quadrilateral with point $L$ as the center of the circumscribed circle.  $(ii)$ $BI_a$ and $CI_a$ bisect the exterior angles of $\angle B$ and $\angle C$,  respectively.  Proof $(i)$ Notice that the the question is equivalent to proving the distance from $L$ to points $B, I, C, I_a$ are equal. In other words, we need to show that  $$LB = LI = LC = LI_a$$ Firstly, it is obvious that $LI = LI_a$ because by definition, $I_a$ is the reflection of $I$ over $L$, and reflections preserve the length of the segment.  Now, we are left with proving that $LB=LI$, because similar calculations will provide $LI=LC$.  We see that most of our given information involves angles, so we focus on proving $\angle LBI = \angle LIB$. Firstly,  $$\angle LBC = \angl...

Percent is another way of saying 'for every hundred'.     Any division where the divisor is $100$ is a percentage.         For example: $ \cfrac {20}{100} = 20 \%$  or                             $ \cfrac {45}{100} = 45 \%$ In essence saying that,                              $ \cfrac {x}{100} = x \%$ Since any ratio can also be expressed as a division, it can also be represented as a percentage.  For ex., a ratio of $\cfrac{1}{2}$ or $1:2$ can be converted to a percent.  $ \cfrac {1}{2}$ $=$ $ \cfrac {1 * 50}{2 * 50} = \cfrac {50}{100} = 50$ Per Cent $=  50\%$  Expressing $x \%$ as a Fraction Any percentage can be expressed as a decimal fraction by dividing the percentage by $100$.  As $x \% = x$ out of $100 = \cfrac {x}{100}$ $75 \% = 75$...