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## Graph Coordinates | AMC 10A, 2015 | Question 12

Try this beautiful Problem on Graph Coordinates from coordinate geometry from AMC 10A, 2015.

## Graph Coordinates – AMC-10A, 2015- Problem 12

Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^{2}+x^{4}=2 x^{2} y+1 .$ What is $|a-b| ?$

,

• $0$
• $1$
• $2$
• $3$
• $4$

### Key Concepts

Co-ordinate geometry

graph

Distance Formula

Answer: $2$

AMC-10A (2015) Problem 12

Pre College Mathematics

## Try with Hints

The given points are $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ which are satisfying the equation $y^{2}+x^{4}=2 x^{2} y+1$.

So we can write $y^{2}+\sqrt{\pi}^{4}=2 \sqrt{\pi}^{2} y+1$

Can you now finish the problem ……….

Therefore

$y^{2}+\pi^{2}=2 \pi y+1$
$y^{2}-2 \pi y+\pi^{2}=1$
$(y-\pi)^{2}=1$
$y-\pi=\pm 1$
$y=\pi+1$
$y=\pi-1$

can you finish the problem……..

There are only two solutions to the equation, so one of them is the value of $a$ and the other is $b$. our requirement is $|a-b|$ so between a and b which is greater is not importent…………

So, $|(\pi+1)-(\pi-1)|=2$

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## Sum of Co-ordinates | AMC-10A, 2014 | Problem 21

Try this beautiful problem from Algebra based on Sum of Co-ordinates

## Sum of Co-ordinates – AMC-10A, 2014- Problem 21

Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$-axis at the same point. What is the sum of all possible $x$-coordinates of these points of intersection?

• $16$
• $12$
• $-8$
• $-4$

### Key Concepts

Geometry

Co-ordinate

Answer: $-8$

AMC-10A (2014) Problem 21

Pre College Mathematics

## Try with Hints

The given equations are $y=ax+5$$\Rightarrow x=\frac{-5}{a}$…..(1)

$y=3x+b$$\Rightarrow x=\frac{-b}{3}$…………(2)

Since two lines intersect the $x$-axis at the same point,then at first we have to find out the common point on x -axis………

Now the intercept form of the given two equations will be

$\frac{x}{(-5/a)} +\frac{y}{5}=1$ ,Therefore the straight line intersect x-axis at the point ($\frac{-5}{a},0$)

$\frac{x}{(-b/3)}+\frac{y}{b}=1$ Therefore the straight line intersect x-axis at the point ($\frac{-b}{3},0)$.

Can you now finish the problem ……….

Since two lines intersect the $x$-axis at the same point so we may say that

($\frac{-5}{a},0$)=($\frac{-b}{3},0)$

$\Rightarrow ab=15$

The only possible pair (a,b) will be $(1,15),(3,5),(5,3),(15,1)$

can you finish the problem……..

Now if we put the values $(1,15),(3,5),(5,3),(15,1)$ in (1) & (2) we will get $-5$,$\frac{-5}{3}$,$-1$,$\frac{1}{3}$

Therefore the sun will be $-8$

Categories

## Good numbers Problem | PRMO-2019 | Problem 12

Try this beautiful problem from PRMO, 2019 based on Good numbers.

## Good numbers Problem | PRMO | Problem-12

A natural number $k >$ is called good if there exist natural numbers
$a_1 < a_2 < ………. < a_k$

$\frac{1}{\sqrt a_1} +\frac{1}{\sqrt a_2}+………………. +\frac{1}{\sqrt a_k}=1$

Let $f(n)$ be the sum of the first $n$ good numbers, $n \geq 1$. Find the sum of all values of $n$ for which
$f(n + 5)/f(n)$ is an integer.

• $20$
• $18$
• $13$

### Key Concepts

Number theory

Good number

Integer

Answer:$18$

PRMO-2019, Problem 12

Pre College Mathematics

## Try with Hints

A number n is called a good number if It is a square free number.

Let $a_1 ={A_1}^2$,$a_2={A_2}^2$,………………$a_k={A_k}^2$
we have to check if it is possible for distinct natural number $A_1, A_2………….A_k$ to satisfy,
$\frac{1}{A_1}+\frac{1}{A_2}+………..+\frac{1}{A_k}=1$

Can you now finish the problem ……….

For $k = 2$; it is obvious that there do not exist distinct$A_1, A_2$, such that $\frac{1}{A_1}+\frac{1}{A_2}=1 \Rightarrow 2$ is not a good number

For $k = 3$; we have $\frac{1}{2} +\frac{1}{3}+\frac{1}{6}=1 \Rightarrow 3$ is a good number.

$\frac{1}{2}+\frac{1}{2}\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ $\Rightarrow 4$ is a good number

Let $k$ wil be a good numbers for all $k \geq 3$

$f(n) = 3 + 4 +… n$ terms =$\frac{n(n + 5)}{2}$
$f(n + 5) =\frac{(n + 5)(n +10)}{2}$

$\frac{f(n+5}{f(n)}=\frac{n+10}{n}=1+\frac{10}{n}$

Can you finish the problem……..

Therefore the integer for n = $1$, $2$, $5$ and $10$. so sum=$1 + 2 + 5 + 10 = 18$.

Categories

## Direction & Angles | PRMO-2019 | Problem 4

Try this beautiful Geometry problem from PRMO, 2019 based on Direction & Angles.

## Direction & Angles | PRMO | Problem-4

An ant leaves the anthill for its morning exercise. It walks 4 feet east and then makes a 160° turn to the right and walks 4 more feet. It then makes another 160° turn to the right and walks 4 more feet. If the ant continues this pattern until it reaches the anthill again, what is the distance in feet it would have walked?

• $20$
• $36$
• $13$

### Key Concepts

Geometry

Co-ordinate geometry

Trigonometry

Answer:$36$

PRMO-2018, Problem 13

Pre College Mathematics

## Try with Hints

According to the problem we draw the figure and try to solve using co-ordinate method…

Can you now finish the problem ……….

Let $A_0(0,0)$

Therefore $A_1(4 cos 0,4sin 0)$

$A_2(4cos 0+4 cos 160, 4 sin 0+4 sin 160)$

$\Rightarrow A_n =(0,0)$

and $4(cos 0+cos 160 +…+sin 160(n-1)=0$

$\Rightarrow n =9$

Can you finish the problem……..

Therefore distance covered =$4 \times 9$=$36$

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## Circles and semi-circles| AMC 8, 2010|Problem 23

Try this beautiful problem from Geometry based on Ratio of the area of circle and semi-circles.

## Area of circles and semi-circles – AMC-8, 2010 – Problem 23

Semicircles POQ and ROS pass through the center O. What is the ratio of the combined areas of the two semicircles to the area of circle O?

• $\frac{1}{2}$
• $\frac{2}{\pi}$
• $\frac{3}{2}$

### Key Concepts

Geometry

Circle

co-ordinate geometry

Answer:$\frac{1}{2}$

AMC-8 (2010) Problem 23

Pre College Mathematics

## Try with Hints

Find the radius of the circle

Can you now finish the problem ……….

Join O and Q

can you finish the problem……..

The co-ordinate of Q is (1,1), So OB=1 and BQ=1

By the Pythagorean Theorem, the radius of the larger circle i.e OQ=$\sqrt{1^2+1^2}$=$\sqrt 2$.

Therefore the area of the larger circle be $\pi (\sqrt 2)^2=2\pi$

Now for the semicircles, radius OB=OC=1(as co-ordinate of P=(1,1) and S=(1,-1))

So, the area of the two semicircles is  $2\times\frac{\pi(1)^2}{2}=\pi$

Finally, the ratio of the combined areas of the two semicircles to the area of circle O is

$\frac{\pi}{2\pi}$=$\frac{1}{2}$

Categories

## Co-ordinate Geometry – AMC 10B – 2019 – Problem No – 4

This problem on co-ordinate geometry is from AMC 10B, 2019. First try it yourself.

All lines with equation ax+by= c such that a,b,c form an arithmetic progression pass through a common point . What are the coordinates of that point?

• (-1,2)
• (0,1)
• (1,-2)
• (1,0)

### Key Concepts

Arithmetic Progression

2D – Co- ordinate Geometry

Cartesian System of Points

AMC 10B – 2019 – Problem No – 4

Challenges and Thrills in Pre-College Mathematics

## Try with Hints

If all lines satisfy the condition, then we can just plug in values for a,b,c that form an arithmetic progression. Let’s use a =1,b= 2,c=3 and a= 1,b=3,c =5. Then the two lines we get are:

x+2y =3

x+3y = 5 so

y= 2

Now plug the value of y into one of the previous equations. We get :

x+4=3

x = -1

Ans is (-1,2)

Categories

# Understand the problem

Let $ABC$ be an equilateral triangle and $P$ in its interior. The distances from $P$ to the triangle’s sides are denoted by $a^2, b^2,c^2$respectively, where $a,b,c>0$. Find the locus of the points $P$ for which $a,b,c$ can be the sides of a non-degenerate triangle.

##### Source of the problem
Romanian Master in Mathematics, 2008
Geometry
##### Difficulty Level
Medium
This problem obviously points towards an application of trilinear coordinates (see more here).

Do you really need a hint? Try it first!

If there exists a non-degenerate triangle with sides $a,b,c$ then the area of the triangle must be positive. Make use of this fact employing Heron’s formula.
Hint 1 will give you a locus in terms of trilinear coordinates. However, it is easier to work in cartesian coordinates as we are already familiar with many curves in them.
Take $A=(1,0,0), B= (0,1,0), C= (0,0,1)$. There is a simple relationship between the cartesian and trilinear coordinates for this choice.

.Note that all points in the plane of $ABC$ satisfy $x+y+z=1$ (why?). For any $P$ in the interior, Let $Q$ be the foot of the perpendicular from $P$ to $AB$ and $R$ be the foot of the perpendicular from $P$ to the XY plane. It is possible to show that $\frac{PR}{PQ}= \sqrt{\frac{2}{3}}$ (it is because the angle between the plane and the Z axis is $\arccos \sqrt{\frac{2}{3}}$). Hence, $\frac{z}{c^2}=\sqrt{\frac{3}{2}}$. By symmetry, $\frac{x}{a^2}=\sqrt{\frac{2}{3}}=\frac{y}{b^2}$. This relates the two coordinate systems. For a triangle with sides $a,b,c$, the square of the area is $\frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$. For this to be positive, we must have (after simplification) $a^4+b^4+c^4< 2(a^2b^2+b^2c^2+c^2a^2)$. In cartesian coordinates, this translates to $\frac{3}{2}(x^2+y^2+z^2)<3(xy+yz+zx)$, which is equivalent to $(x+y+z)^2>2(x^2+y^2+z^2)$. As $P$ lies on the plane $x+y+z=1$, this means that $x^2+y^2+z^2<\frac{1}{2}$. This last equation is that of the interior of a solid sphere. Hence, our desired locus is the intersection of this solid sphere with the plane $x+y+z=1$, which is precisely the interior of the circumcircle of $ABC$.

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