Categories

## Problem on Curve | AMC 10A, 2018 | Problem 21

Try this beautiful Problem on Algebra based on Problem on Curve from AMC 10 A, 2018. You may use sequential hints to solve the problem.

## Curve- AMC 10A, 2018- Problem 21

Which of the following describes the set of values of $a$ for which the curves $x^{2}+y^{2}=a^{2}$ and $y=x^{2}-a$ in the real $x y$ -plane intersect at
exactly 3 points?

• $a=\frac{1}{4}$
• $\frac{1}{4}<a<\frac{1}{2}$
• $a>\frac{1}{4}$
• $a=\frac{1}{2}$
• $a>\frac{1}{2}$

Algebra

greatest integer

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2018 Problem-14

#### Check the answer here, but try the problem first

$a>\frac{1}{2}$

## Try with Hints

#### First Hint

We have to find out the value of $a$

Given that $y=x^{2}-a$ . now if we Substitute this value in $x^{2}+y^{2}=a^{2}$ we will get a quadratic equation of $x$ and $a$. if you solve this equation you will get the value of $a$

Now can you finish the problem?

#### Second Hint

After substituting we will get $x^{2}+\left(x^{2}-a\right)^{2}$=$a^{2} \Longrightarrow x^{2}+x^{4}-2 a x^{2}=0 \Longrightarrow x^{2}\left(x^{2}-(2 a-1)\right)=0$

therefore we can say that either $x^2=0\Rightarrow x=0$ or $x^2-(2a-1)=0$

$\Rightarrow x=\pm \sqrt {2a-1}$. Therefore

Now Can you finish the Problem?

#### Third Hint

Therefore $\sqrt {2a-1} > 0$

$\Rightarrow a>\frac{1}{2}$

Categories

## ISI MStat PSB 2006 Problem 2 | Cauchy & Schwarz come to rescue

This is a very subtle sample problem from ISI MStat PSB 2006 Problem 2. After seeing this problem, one may think of using Lagrange Multipliers, but one can just find easier and beautiful way, if one is really keen to find one. Can you!

## Problem– ISI MStat PSB 2006 Problem 2

Maximize $x+y$ subject to the condition that $2x^2+3y^2 \le 1$.

### Prerequisites

Cauchy-Schwarz Inequality

Tangent-Normal

Conic section

## Solution :

This is a beautiful problem, but only if one notices the trick, or else things gets ugly.

Now we need to find the maximum of $x+y$ when it is given that $2x^2+3y^2 \le 1$. Seeing the given condition we always think of using Lagrange Multipliers, but I find that thing very nasty, and always find ways to avoid it.

So let’s recall the famous Cauchy-Schwarz Inequality, $(ab+cd)^2 \le (a^2+c^2)(b^2+d^2)$.

Now, lets take $a=\sqrt{2}x ; b=\frac{1}{\sqrt{2}} ; c= \sqrt{3}y ; d= \frac{1}{\sqrt{3}}$, and observe our inequality reduces to,

$(x+y)^2 \le (2x^2+3y^2)(\frac{1}{2}+\frac{1}{3}) \le (\frac{1}{2}+\frac{1}{3})=\frac{5}{6} \Rightarrow x+y \le \sqrt{\frac{5}{6}}$. Hence the maximum of $x+y$ with respect to the given condition $2x^2+3y^2 \le 1$ is $\frac{5}{6}$. Hence we got what we want without even doing any nasty calculations.

Another nice approach for doing this problem is looking through the pictures. Given the condition $2x^2+3y^2 \le 1$ represents a disc whose shape is elliptical, and $x+y=k$ is a family of straight parallel lines passing passing through that disc.

Hence the line with the maximum intercept among all the lines passing through the given disc represents the maximized value of $x+y$. So, basically if a line of form $x+y=k_o$ (say), is a tangent to the disc, then it will basically represent the line with maximum intercept from the mentioned family of line. So, we just need to find the point on the boundary of the disc, where the line of form $x+y=k_o$ touches as a tangent. Can you finish the rest and verify weather the maximum intercept .i.e. $k_o= \sqrt{\frac{5}{6}}$ or not.

## Food For Thought

Can you show another alternate solution to this problem ? No, Lagrange Multiplier Please !! How would you like to find out the point of tangency if the disc was circular ? Show us the solution we will post them in the comment.

Keep thinking !!

Categories

## ISI MStat 2019 PSA Problem 14 | Reflection of a point

This is a problem from ISI MStat 2019 PSA Problem 14. First, try the problem yourself, then go through the sequential hints we provide.

## Reflection of a point – ISI MStat Year 2019 PSA Question 14

The reflection of the point (1,2) with respect to the line $x+2 y=15$ is

• (3,6)
• (6,3)
• (10,5)
• (5,10)

### Key Concepts

Straight line

ISI MStat 2019 PSA Problem 14

Precollege Mathematics

## Try with Hints

Find an algorithm to find the reflection,

Find the line perpendicular to $x+2 y=15$ through (1,2).
Find the point of intersection.
Use Midpoint Segment Result.

The line perpendicular to $x+2 y=15$ is of the form $-2x+y=k$ .Now it passes through (1,2) . So, $-2+2=k \Rightarrow k=0$

Hence the line perpendicular to $x+2 y=15$ through (1,2) is y=2x.

Now we will find point of intersection (Foot of Perpendicular )

(3,6) is the point of intersection i.e the foot of perpendicular.

Use Mid-Point Formula (special case of Section formula) to get required point (Foot of perpendicular is mid-point of reflection and original point)

$(3,6)=( \frac{x+1}{2} , \frac{y+2}{2} )$ $\Rightarrow x=5 , y=10$

Therefore the reflection point is (5,10) .

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## ISI MStat 2016 PSA Problem 9 | Equation of a circle

This is a problem from ISI MStat 2016 PSA Problem 9 based on equation of a circle. First, try the problem yourself, then go through the sequential hints we provide.

## Equation of a circle- ISI MStat Year 2016 PSA Question 9

Given $\theta$ in the range $0 \leq \theta<\pi,$ the equation $2 x^{2}+2 y^{2}+4 x \cos \theta+8 y \sin \theta+5=0$ represents a circle for all $\theta$ in the interval

• $0 < \theta <\frac{\pi}{3}$
• $\frac{\pi}{4} < \theta <\frac{3\pi}{4}$
• $0 < \theta <\frac{\pi}{2}$
• $0 \le \theta <\frac{\pi}{2}$

### Key Concepts

Equation of a circle

Trigonometry

Basic Inequality

Answer: is $\frac{\pi}{4} < \theta <\frac{3\pi}{4}$

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

## Try with Hints

Complete the Square.

We get ,

$2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3))$
$6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2}$

We are given that $0 \leq \theta<\pi,$ . So, ${\sin^2 \theta} \geq \frac{1}{2}$ $\Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4}$.

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## Sides of Quadrilateral | PRMO 2017 | Question 20

Try this beautiful problem from the Pre-RMO, 2017 based on Sides of Quadrilateral.

## Sides of Quadrilateral – PRMO 2017

What is the number of triples (a,b,c) of positive integers such that (i) a<b<c<10 and (ii) a,b,c,10 form the sides of a quadrilateral?

• is 107
• is 73
• is 840
• cannot be determined from the given information

### Key Concepts

Largest number of triples

Distance

PRMO, 2017, Question 20

Geometry Vol I to IV by Hall and Stevens

First hint

a+b+c>10

(a+b+c) can be

a b c

1 2 8,9

1 3 7,8,9

1 4 6 ,7,8,9

1 5 6,7,8,9

1 6 7,8,9

1 7 8,9

1 8 9

Second Hint

2 3 6,7,8,9

2 4 5,6,7,8,9

2 5 6,7,8,9

2 6 7,8,9

2 7 8,9

2 8 9

Final Step

3 4 5,6,7,8,9

3 5 6,7,8,9

3 6 7,8,9

3 7 8,9

3 8 9

4 5 6,7,8,9

4 6 7,8,9

4 7 8,9

4 8 9

5 6 7,8,9

5 7 8,9

5 8 9

6 7 8,9

6 8 9

7 8 9

Total 73 cases.

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## Ordered pair Problem | AIME I, 1987 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Ordered pair.

## Ordered pair Problem – AIME I, 1987

An ordered pair (m,n) of non-negative integers is called simple if the additive m+n in base 10 requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to 1492.

• is 107
• is 300
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Ordered pair

Algebra

AIME I, 1987, Question 1

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

for no carrying required

Second Hint

the range of possible values of any digit m is from 0 to 1492 where the value of n is fixed

Final Step

Number of ordered pair (1+1)(4+1)(9+1)(2+1)

=(2)(5)(10)(3)

=300.

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## Hyperbola & Tangent | ISI MStat 2016 Problem 1 | PSB Sample

This is a beautiful sample problem from ISI MStat 2016 PSB Problem 1. This is based on finding the minimum value of a function subjected to the restriction.

## ISI MStat 2016 Problem 1

Let $x, y$ be real numbers such that $x y=10$ . Find the minimum value of $|x+y|$ and all also find all the points $(x, y)$ where this minimum value is achieved.

## Solution

(a) Using graph

The equation $xy=10={(\sqrt{10})}^2$ represents the equation of rectangular hyperbola with foci are $(- \sqrt{10},- \sqrt{10})$ and $(\sqrt{10}, \sqrt{10} )$ .

Now , $|x+y|=c \Rightarrow$ $x+y=\pm c$ , which looks somewhat like this ,

we have to find the the minimum value of $|x + y|$ subject to the restriction that $xy=10$ . If we move $|x+y|=c$ along $xy =10$ by varying c , then local minimum can occur at the points where the level curve $|x+y|=c$ touch $xy =10$ . Now as both the rectangular hyperbola and |x+y|=c are symmetric about $x+y=0$ for $c \ne 0$ and $x=y$ , the level curve will touch $xy =10$ when x+y=c and x+y=-c both are tangent to the curve $xy =10$ . And tangents occurs at the foci of $xy =10$ i.e at $(- \sqrt{10},- \sqrt{10})$ and $(\sqrt{10}, \sqrt{10} )$ .

Hence , the minimum value of $|x + y|$ are $|\sqrt{10} +\sqrt {10} |$ and $|-\sqrt{10}- \sqrt{10}|$ both gives the same value $2 \sqrt{10}$ .

Therefore , the minimum value of $|x + y|$ is $2 \sqrt{10}$ and it attains it’s minimum at $(\sqrt{10} , \sqrt{10} )$ and $( – \sqrt{10} , -\sqrt{10})$ .

(b) Using Derivative test

$|x+y|= |x+ \frac{10}{x} |$ as we are given that $xy=10$

Let, $f(x) =|x+ \frac{10}{x}|$ then we have to find the minimum value of f(x)

Now ,as the function is not defined at x=0 and also x=0 can’t give the minimum value of |x+y| due to the condition that xy=10. So, we will study f(x) for two cases when x>0 and when x<0 .

$f(x) = \begin{cases} x+ \frac{10}{x} & ,x > 0 \\ -(x+ \frac{10}{x} ) & ,x < 0 \end{cases}$

$f'(x)=\begin{cases} 1- \frac{10}{x^2} & ,x > 0 \\ -(1- \frac{10}{x^2} ) & ,x < 0 \end{cases}$

$f'(x)=0 \Rightarrow$ $x= \pm \sqrt{10}$

$f”(x)= \begin{cases} \frac{20}{x^3} & ,x > 0 \\ -\frac{20}{x^3} & ,x < 0 \end{cases}$

So,$f”(x) >0$ for $x= \pm \sqrt{10}$

Hence f(x) attains it’s minimum value at $(\sqrt{10} , \sqrt{10} )$ and $– \sqrt{10} , -\sqrt{10})$ and minimum value is $2 \sqrt{10}$.

## Challenge Problem

Let $x_1, x_2 ,…, x_n$ be be positive real numbers such that $\prod_{i=1}^{n} x_{i} = 10$ . Find the minimum value of $\sum_{i=1}^{n} x_{i}$.

Categories

## GCD and Ordered pair | AIME I, 1995 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on GCD and Ordered pair.

## GCD and Ordered pair – AIME I, 1995

Find number of ordered pairs of positive integers (x,y) with $y \lt x \leq 100$ are both $\frac{x}{y}$ and $\frac{x+1}{y+1}$ integers.

• is 107
• is 85
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

GCD

Ordered pair

AIME I, 1995, Question 8

Elementary Number Theory by David Burton

## Try with Hints

First hint

here y|x and (y+1)|(x+1) $\Rightarrow gcd(y,x)=y, gcd(y+1,x+1)=y+1$

$\Rightarrow gcd(y,x-y)=y, gcd(y+1,x-y)=y+1$

$\Rightarrow y,y+1|(x-y) and gcd (y,y+1)=1$

$\Rightarrow y(y+1)|(x-y)$

Second Hint

here number of multiples of y(y+1) from 0 to 100-y $(x \leq 100)$ are

[$\frac{100-y}{y(y+1)}$]

Final Step

$\Rightarrow \displaystyle\sum_{y=1}^{99}[\frac{100-y}{y(y+1)}$]=49+16+8+4+3+2+1+1+1=85.

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## Coordinate Geometry Problem | AIME I, 2009 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Coordinate Geometry.

## Coordinate Geometry Problem – AIME 2009

Consider the set of all triangles OPQ where O  is the origin and P and Q are distinct points in the plane with non negative integer coordinates (x,y) such that 41x+y=2009 . Find the number of such distinct triangles whose area is a positive integer.

• is 107
• is 600
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Geometry

AIME, 2009, Question 11

Geometry Revisited by Coxeter

## Try with Hints

First hint

let P and Q be defined with coordinates; P=($x_1,y_1)$ and Q($x_2,y_2)$. Let the line 41x+y=2009 intersect the x-axis at X and the y-axis at Y . X (49,0) , and Y(0,2009). such that there are 50 points.

here [OPQ]=[OYX]-[OXQ] OY=2009 OX=49 such that [OYX]=$\frac{1}{2}$OY.OX=$\frac{1}{2}$2009.49 And [OYP]=$\frac{1}{2}$$2009x_1$  and [OXQ]=$\frac{1}{2}$(49)$y_2$.

Second Hint

2009.49 is odd, area OYX not integer of form k+$\frac{1}{2}$ where k is an integer

Final Step

41x+y=2009 taking both 25  $\frac{25!}{2!23!}+\frac{25!}{2!23!}$=300+300=600.

.