Categories

## Arithmetic Mean | AIME I, 2015 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Arithmetic Mean.

## Arithmetic Mean of Number Theory – AIME 2015

Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

• is 107
• is 431
• is 840
• cannot be determined from the given information

### Key Concepts

Inequalities

Algebra

Number Theory

AIME, 2015, Question 12

Elementary Number Theory by David Burton

## Try with Hints

Each 1000-element subset ${ a_1, a_2,a_3,…,a_{1000}}$ of ${1,2,3,…,2015}$ with $a_1<a_2<a_3<…<a_{1000}$ contributes $a_1$ to sum of least element of each subset and set ${a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$. $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<…<a_{1000}+1$ ($k$ can be anything from $1$ to $a_1$ inclusive

Thus, the number of ways to choose the set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is equal to the sum. But choosing a set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is same as choosing a 1001-element subset from ${1,2,3,…,2016}$!

average =$\frac{2016}{1001}$=$\frac{288}{143}$. Then $p+q=288+143={431}$

Categories

## Ratio and Inequalities | AIME I, 1992 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Ratio and Inequalities.

## Ratio and Inequalities – AIME I, 1992

A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly 0.500. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than 0.503. Find the largest number of matches she could have won before the weekend began.

• is 107
• is 164
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Ratios

Inequalities

AIME I, 1992, Question 3

Elementary Algebra by Hall and Knight

## Try with Hints

Let x be number of matches she has played and won then $\frac{x}{2x}=\frac{1}{2}$

and $\frac{x+3}{2x+4}>\frac{503}{1000}$

$\Rightarrow 1000x+3000 > 1006x+2012$

$\Rightarrow x<\frac{988}{6}$

$\Rightarrow$ x=164.

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## Length and Inequalities | AIME I, 1994 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Length and Inequalities.

## Length and Inequalities – AIME I, 1994

A fenced, rectangular field measures 24 meters by 52 meters.An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the field, find the largest number of square test plots into which the field can be partitioned using all or some of the 1994 meters of fence.

• is 107
• is 702
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Inequalities

Length

AIME I, 1994, Question 12

Inequalities (Little Mathematical Library) by Korovkin

## Try with Hints

Number of squares in a row=$\frac{52n}{24}$=$\frac{13n}{6}$ squares in every row

each vertical fence lengths 24 for $\frac{13n}{6}-1$ vertical fences

each horizontal fence lengths 52 for n-1 such fences

total length of internal fencing 24 ($\frac{13n}{6}-1$)+52(n-1)=104n-76 $\leq 1994$

$\Rightarrow n \leq \frac{1035}{52}$

$\Rightarrow n \leq 19$

the largest multiple of 6 that is $\leq 19$

$\Rightarrow n=18$

required number =$\frac{13n^{2}}{6}$=702.

Categories

## Perfect square Problem | AIME I, 1999 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Perfect square and Integers.

## Perfect square Problem – AIME I, 1999

Find the sum of all positive integers n for which $n^{2}-19n+99$ is a perfect square.

• is 107
• is 38
• is 840
• cannot be determined from the given information

### Key Concepts

Perfect Square

Integers

Inequalities

AIME I, 1999, Question 3

Elementary Number Theory by David Burton

## Try with Hints

$(n-10)^{2}$ $\lt$ $n^{2}-19n+99$ $\lt$ $(n-8)^{2}$ and $n^{2}-19n+99$ is perfect square then $n^{2}-19n+99$=$(n-9)^{2}$ that is n=18

and $n^{2}-19n+99$ also perfect square for n=1,9,10

Categories

## Sequence and Integers | AIME I, 2007 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2007 based on Sequence and Integers.

## Sequence and Integers – AIME I, 2007

A sequence is defined over non negetive integral indexes in the following way $a_0=a_1=3$, $a_{n+1}a_{n-1}=a_n^{2}+2007$, find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$

• is 107
• is 224
• is 840
• cannot be determined from the given information

### Key Concepts

Sequence

Inequalities

Integers

AIME I, 2007, Question 14

Elementary Number Theory by David Burton

## Try with Hints

$a_{n+1}a_{n-1}$=$a_{n}^{2}+2007$ then $a_{n-1}^{2} +2007 =a_{n}a_{n-2}$ adding these $\frac{a_{n-1}+a_{n+1}}{a_{n}}$=$\frac{a_{n}+a_{n-2}}{a_{n-1}}$, let $b_{j}$=$\frac{a_{j}}{a_{j-1}}$ then $b_{n+1} + \frac{1}{b_{n}}$=$b_{n}+\frac{1}{b_{n-1}}$ then $b_{2007} + \frac{1}{b_{2006}}$=$b_{3}+\frac{1}{b_{2}}$=225

here $\frac{a_{2007}a_{2005}}{a_{2006}a_{2005}}$=$\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$ then $b_{2007}$=$\frac{a_{2007}}{a_{2006}}$=$\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$$\gt$$\frac{a_{2006}}{a_{2005}}$=$b_{2006}$

then $b_{2007}+\frac{1}{b_{2007}} \lt b_{2007}+\frac{1}{b_{2006}}$=225 which is small less such that all $b_{j}$ s are greater than 1 then $\frac{a_{2006}^{2}+ a_{2007}^{2}}{a_{2006}a_{2007}}$=$b_{2007}+\frac{1}{b_{2007}}$=224.

Categories

## Smallest Perimeter of Triangle | AIME I, 2015 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Smallest Perimeter of Triangle.

## Smallest Perimeter of Triangle – AIME 2015

Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\triangle ABC$..

• is 107
• is 108
• is 840
• cannot be determined from the given information

### Key Concepts

Inequalities

Trigonometry

Geometry

AIME, 2015, Question 11

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

Let $D$ be the midpoint of $\overline{BC}$. Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$, so $\angle ADB = \angle ADC = 90^o$.Now let $BD=y$, $AB=x$, and $\angle IBD$ =$\frac{\angle ABD}{2}$ = $\theta$.Then $\mathrm{cos}{(\theta)} = \frac{y}{8}$and $\mathrm{cos}{(2\theta)} = \frac{y}{x} = 2\mathrm{cos^2}{(\theta)} – 1 = \frac{y^2-32}{32}$.

Cross-multiplying yields $32y = x(y^2-32)$.

Since $x,y>0$, $y^2-32$ must be positive, so $y > 5.5$.

Additionally, since $\triangle IBD$ has hypotenuse $\overline{IB}$ of length $8$, $BD=y < 8$.

Therefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$, $6.5$, $7$, and $7.5$.

However, only one of these values, $y=6$, yields an integral value for $AB=x$, so we conclude that $y=6$ and $x=\frac{32(6)}{(6)^2-32}=48$.

Thus the perimeter of $\triangle ABC$ must be $2(x+y) = {108}$.

Categories

## Triangle Inequality Problem – AMC 12B, 2014 – Problem 13

Try this beautiful problem from AMC 12 based on Triangle inequality problem.

## Problem – Triangle Inequality

Real numbers a and b are chosen with 1 < a < b such that no triangle with positive area has side lengths 1,a and b or $\frac {1}{b},\frac {1}{a}$ and 1. What is the smallest possible value of b?

• $\frac {3+\sqrt 3}{2}$
• $\frac {5}{2}$
• $\frac {3+\sqrt 5}{2}$
• $\frac {3+\sqrt 6}{2}$

### Key Concepts

Triangle Inequality

Inequality

Geometry

Answer: $\frac {3+\sqrt 5}{2}$

American Mathematics Competition – 12B ,2014, Problem Number – 13

Secrets in Inequalities.

## Try with Hints

Here is the first hint where you can start this sum:

It is given $1 >\frac {1}{a} > \frac {1}{b }$ . Use Triangle Inequality here :

a+1>b

a>b-1

$\frac {1}{a} + \frac {1}{b} >1$

If we want to find the lowest possible value of b , we create we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. Thus we get : a = b – 1

Now try to do the rest of the sum.………………….

We already know $\frac {1}{a} + \frac {1}{b} = 1$

After substituting we will get :

$\frac {1}{b – 1} + \frac {1}{b} = \frac {b+b-1}{b(b-1)} = 1$

$\frac {2b – 1}{b(b-1)} = 1$

Now do the rest of the calculation ………………………..

Here is the rest of steps to check your problem :

$2b – 1 = b^2 – b$

Now Solving for b using the quadratic equation, we get

$b^2 – 3b + 1 = 0$

$b = \frac {3 + \sqrt 5}{2}$ (Answer)

Categories

## Problem – Basic Inequality

To enter a junior writing competition you must be under 12 years old. Which of the following is correct?

• Age < 12
• Age ≤ 12
• Age > 12
• Age ≥ 12

### Key Concepts

Inequality

Mathematical analysis

Numbers

Basic Problems of Inequalities

Secrets in Inequalities.

## Try with Hints

The basic rule for inequalities can be explained by some imaginary values ………..

Its one of the easiest sum. Here we only have to find the relation between the perimeter of this sum.We have two different perimeter

1. Age
2. Limitation on age
3. So can you try to find the relation between these two that mentioned in this sum. Try to find it …………………………………….

For this sum its mentioned that the age is under 12 that means the age is less than 12.

If we try to implement it in this sum we can express it as :

Age $\leq 12$