Categories

## Finding side of Triangle | PRMO-2014 | Problem 15

Try this beautiful problem from PRMO, 2014 based on Finding side of Triangle.

## Finding side of Triangle | PRMO | Problem 15

Let XOY be a triangle with angle XOY=90 degrees. Let M and N be the midpoints of the legs OX and OY, respectively. Suppose that XN=19 and YM=22. what is XY?

• $28$
• $26$
• $30$

### Key Concepts

Geometry

Triangle

Pythagoras

Answer:$26$

PRMO-2014, Problem 15

Pre College Mathematics

## Try with Hints

Given that $\angle XOY=90^{\circ}$ .Let M and N be the midpoints of the legs OX and OY, respectively, and that XN=19 and YM=22. Now $\triangle XON$ & $\triangle MOY$ are Right angle Triangle. Use Pythagoras theorem …….

Can you now finish the problem ……….

Let $XM=MO=p$ and $ON=NY=q$.Now using Pythagoras theorm on $\triangle XON$ & $\triangle MOY$ we have…

$OX^2 +ON^2=XN^2$ $\Rightarrow 4p^2 +q^2=19^2$ $\Rightarrow 4p^2 +q^2=361$………..(1) and $OM^2 +OY^2=MY^2$ $\Rightarrow p^2 +4q^2=22^2$ $\Rightarrow p^2 +4q^2=484$……(2)

Now Adding (1)+(2)=$(4p^2 +q^2=361)$+$(p^2 +4q^2=484$ $\Rightarrow 5(p^2+q^2)=845$ $\Rightarrow (p^2+q^2)=169$ $\Rightarrow 4(p^2+q^2)=676$ $\Rightarrow (OX)^2+(OY)^2=(26)^2$ $\Rightarrow (XY)^2=(26)^2$ $\Rightarrow XY=26$.

Categories

## Ordered Pairs | PRMO-2019 | Problem 18

Try this beautiful problem from PRMO, 2019, Problem 18 based on Ordered Pairs.

## Orderd Pairs | PRMO | Problem-18

How many ordered pairs $(a, b)$ of positive integers with $a < b$ and $100 \leq a$, $b \leq 1000$ satisfy $gcd (a, b) : lcm (a, b) = 1 : 495$ ?

• $20$
• $91$
• $13$
• $23$

### Key Concepts

Number theory

Orderd Pair

LCM

Answer:$20$

PRMO-2019, Problem 18

Pre College Mathematics

## Try with Hints

At first we assume that $a = xp$
$b = xq$
where $p$ & $q$ are co-prime

Therefore ,

$\frac{gcd(a,b)}{LCM(a ,b)} =\frac{495}{1}$

$\Rightarrow pq=495$
Can you now finish the problem ……….

Therefore we can say that

$pq = 5 \times 9 \times 11$
$p < q$

when $5 < 99$ (for $x = 20, a = 100, b = 1980 > 100$),No solution
when $9 < 55$ $(x = 12$ to $x = 18)$,7 solution
when,$11 < 45$ $(x = 10$ to $x = 22)$,13 solution
Can you finish the problem……..

Therefore Total solutions = $13 + 7=20$

Categories

## Maximum area | PRMO-2019 | Problem 23

Try this beautiful problem from PRMO, 2019 based on Maximum area

## Maximum area | PRMO-2019 | Problem-23

Let $\mathrm{ABCD}$ be a convex cyclic quadrilateral. Suppose $\mathrm{P}$ is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least. If ${\mathrm{PA}, \mathrm{PB}, \mathrm{PC}, \mathrm{PD}}={3,4,6,8} .$ What is the maximum possible area of ABCD?

• $20$
• $55$
• $13$
• $23$

### Key Concepts

Geometry

Triangle

Area

Answer:$55$

PRMO-2019, Problem 23

Pre College Mathematics

## Try with Hints

Given that $\mathrm{PA}=\mathrm{a}, \mathrm{PB}=\mathrm{b}, \mathrm{PC}=\mathrm{c}, \mathrm{PD}=\mathrm{d}$
Now from the above picture area of quadrilateral ABCD
Area=$[\mathrm{APB}]+[\mathrm{BPC}]+[\mathrm{CPD}]+[\mathrm{DPA}]$

Therefore area $\Delta=\frac{1}{2} \mathrm{ab} \sin \mathrm{x}+\frac{1}{2} \mathrm{bc} \sin \mathrm{y}+\frac{1}{2} \mathrm{cd} \sin \mathrm{z}+\frac{1}{2}$ da $\sin \mathrm{w}$
$\Delta_{\max }$ when $x=y=z=w=90^{\circ}$
$\Delta_{\max }=\frac{1}{2}(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{d})$
Now ac $=$ bd (cyclic quadrilateral) As $(a, b, c, d)=(3,4,6,8)$
$\Rightarrow{(a, c)(b, d)}={(3,8)(4,6)}$

So $\Delta_{\max }=\frac{1}{2} \times 11 \times 10=55$

Categories

## Ratio of the areas | PRMO-2019 | Problem 19

Try this beautiful problem from PRMO, 2019 based on Ratio of the areas.

## Ratio of the areas | PRMO | Problem-19

Let $\mathrm{AB}$ be a diameter of a circle and let $\mathrm{C}$ be a point on the segment $\mathrm{AB}$ such that $\mathrm{AC}: \mathrm{CB}=6: 7 .$ Let $\mathrm{D}$ be a point on the circle such that $\mathrm{DC}$ is perpendicular to $\mathrm{AB}$. Let DE be the diameter through $\mathrm{D}$. If $[\mathrm{XYZ}]$ denotes the area of the triangle XYZ. Find [ABD] / $[\mathrm{CDE}]$ to the nearest integer.

• $20$
• $91$
• $13$
• $23$

### Key Concepts

Geometry

Triangle

Area

Answer:$13$

PRMO-2019, Problem 19

Pre College Mathematics

## Try with Hints

$\angle \mathrm{AOC} \quad=\frac{6 \pi}{13}, \angle \mathrm{BOC}=\frac{7 \pi}{13}$

$\mathrm{Ar} \Delta \mathrm{ABD}=\mathrm{Ar} \Delta \mathrm{ABC}=\frac{1}{2} \mathrm{AB} \times \mathrm{OC} \sin \frac{6 \pi}{13}$

$\mathrm{Ar} \Delta \mathrm{CDE}=\frac{1}{2} \mathrm{DE} \times \mathrm{OC} \sin \left(\frac{7 \pi}{13}-\frac{6 \pi}{13}\right)$

$\frac{[\mathrm{ABD}]}{[\mathrm{CDE}]}=\frac{\sin \frac{6 \pi}{13}}{\sin \frac{\pi}{13}}=\frac{1}{2 \sin \frac{\pi}{26}}=\mathrm{p}$

because $\sin \theta \cong \theta$ if $\theta$ is small
$\Rightarrow \sin \frac{\pi}{26} \cong \frac{\pi}{26}$

$\mathrm{p}=\frac{13}{\pi} \Rightarrow$ Nearest integer to $\mathrm{p}$ is 4

Categories

## Problem on Positive Integers | PRMO-2019 | Problem 26

Try this beautiful problem from Algebra PRMO 2019 based on Positive Integers.

## Positive Integers | PRMO | Problem 26

Positive integers x,y,z satisfy xy+z=160 compute smallest possible value of x+yz.

• 24
• 50
• 29
• 34

### Key Concepts

Algebra

Integer

sum

PRMO-2019, Problem 26

Higher Algebra by Hall and Knight

## Try with Hints

x+yz=$\frac{160-z}{y}$+yz

=$\frac{160}{y}+\frac{z(y^{2}-1)}{y}=\frac{160-z}{y}+\frac{zy^{2}}{y}=\frac{160-z}{y}+zy$

for particular value of z, $x+yz \geq 2\sqrt{z(160-z)}$

or, least value=$2\sqrt{z(160-z)}$ but an integer also

for least value z is also

case I z=1, $x+yz=\frac{159}{y}+y$ or, min value at y=3 which is 56

case II z=2, $x+yz=\frac{158}{y}+2y$ or, min value at y =2 which is 83 (not taken)

case III z=3, $x+yz=\frac{157}{y}+3y$ or, min value at y=1 which is 160 (not taken)

case IV z=4, $x+yz=\frac{156}{y}+4y$ or, min at y=6 which is 50 (taken)

case V z=5, $x+yz=\frac{155}{y}+5y$ or, minimum value at y=5 which is 56 (not taken)

case VI z=6, $x+yz=\frac{154}{y}+6y$ $\geq 2\sqrt{924}$>50

smallest possible value =50.

Categories

## Smallest positive value | Algebra | PRMO-2019 | Problem 13

Try this beautiful problem from Algebra PRMO 2019 based on smallest positive value

## Smallest positive value| PRMO | Problem 13

Each of the numbers $x_1, x_2,……….x_{101}$ is $±1$. What is the smallest positive value of $\displaystyle\sum_{1\leq i \leq j \leq {101}} x_i x_j$ ?

• $24$
• $10$
• $34$

### Key Concepts

Algebra

Integer

sum

Answer:$10$

PRMO-2019, Problem 13

Pre College Mathematics

## Try with Hints

$S=\displaystyle\sum_{1\leq i \leq j \leq {101}} x_i x_j$ .

we have $(x_1+x_2+x_3+….+x_{101})^2={x_1}^2+{x_2}^2+…..+{x_{101}}^2+2S$

$\Rightarrow 2S$=$(\displaystyle\sum_{i=1}^{101} {x_i})^2-\displaystyle\sum_{j=1}^{101} {x_j}^2$

Can you now finish the problem ……….

Since we have $x_i=\pm 1$ so ${x_i}^2=1$

so $2S=(\displaystyle\sum_{i=1}^{101} {x_i})^2-{101}$

Since $\displaystyle\sum_{i=1}^{101} {x_i}$ will be an integer

so $(\displaystyle\sum_{i=1}^{101} {x_i})^2$ will be a perfect square .

For smalll positive $S$, $(\displaystyle\sum_{i=1}^{101} {x_i})^2$must be smallest perfect square greater than ${101}$

So $(\displaystyle\sum_{i=1}^{101} {x_i})^2={121}$

$\Rightarrow (\displaystyle\sum_{i=1}^{101} {x_i})$=${11}$ or ${-11}$

Can you finish the problem……..

We can verify that the desired sum can be achieved by putting $45$ $x_i$’s to be –1 and $56$ $x_i$’s to be $1$ So, $2S = 121 – 101 = 20$

$\Rightarrow s=10$

Categories

## Regular polygon | Combinatorics | PRMO-2019 | Problem 15

Try this beautiful problem from combinatorics PRMO 2019 based on Regular polygon

## Regular polygon| PRMO | Problem 15

In how many ways can a pair of parallel diagonals of a regular polygon of $10$ sides be selected

• $24$
• $45$
• $34$

### Key Concepts

Combinatorics

Regular polygon

geometry

Answer:$45$

PRMO-2019, Problem 15

Pre College Mathematics

## Try with Hints

The above diagram is a diagram of Regular Polygon .we have to draw the diagonals as shown in above.we joined the diagonals such that all the diagonals will be parallel

Can you now finish the problem ……….

If we joined the diagonals (shown in Fig. 1), i.e $(P_3 \to P_10)$,$(P_4\to P_9)$,$(P_5 \to P_8)$ then then we have 3 diagonals.so we have 5$4 \choose 2$ ways=$15$ ways.

If we joined the diagonals (shown in Fig.2), i.e $(P_1 \to P_3)$,$(P_10\to P_4)$,$(P_9\to P_5)$,$(P_8\to P_6)$then we have $4$diagonals.so we have 5$3 \choose 2$ ways=$30$ ways.

Therefore total numbers of ways that can a pair of parallel diagonals of a regular polygon of $10$ sides be selected is $15+30=45$

Categories

## Problem on Real numbers | Algebra | PRMO-2017 | Problem 18

Try this beautiful problem from Algebra PRMO 2017 based on real numbers.

## Problem on Real numbers | PRMO | Problem 18

If the real numbers $x$, $y$, $z$ are such that $x^2 + 4y^2 + 16z^2 = 48$ and $xy + 4yz + 2zx = 24$. what is the
value of $x^2 + y^2 + z^2$ ?

• $24$
• $21$
• $34$

### Key Concepts

Algebra

Equation

Answer:$21$

PRMO-2017, Problem 18

Pre College Mathematics

## Try with Hints

The given equation are

$x^2 + 4y^2 + 16z^2 = 48$
$\Rightarrow (x)2 + (2y)2 + (4z)2 = 48$
$2xy + 8yz + 4zx = 48$
adding tis equations we have to solve the problem….

Can you now finish the problem ……….

Now we can say that
$(x)^2 + (2y)^2 + (4z)^2 – (2xy) – (8yz) – (4zx) = 0$
$\Rightarrow [(x – 2y)2 + (2y – 4z)2 + (x – 4y)2)] = 0$
$x = 2y = 4z$

$\Rightarrow \frac{x}{4}=\frac{y}{2}=z$

Can you finish the problem……..

Therefore we may say that,

$(x, y, z) = (4m, 2m, m)$
$x^2 + 4y^2 + 16z^2 = 48$

$16m^2 + 16m^2 + 16m^2 = 48$
so $m^2 = 1$
$x^2 + y^2 + z^2 = 21m^2 = 21$

Categories

## Roots of Equations | PRMO-2016 | Problem 8

Try this beautiful problem from Algebra based on roots of equations.

## Roots of Equations | PRMO | Problem 8

Suppose that $a$ and $b$ are real numbers such that $ab \neq 1$ and the equations $120 a^2 -120a+1=0$ and $b^2-120b+120=0$ hold. Find the value of $\frac{1+b+ab}{a}$

• $200$
• $240$
• $300$

### Key Concepts

Algebra

Roots

Answer:$240$

PRMO-2016, Problem 8

Pre College Mathematics

## Try with Hints

The given equations are $120 a^2 -120a+1=0$ and $b^2-120b+120=0$.we have to find out the values of $a$ and $b$….

Let $x,y$ be the roots of the equation $120 a^2 -120a+1=0$then $\frac{1}{x},\frac{1}{y}$ be the roots of the equations of $b^2-120b+120=0$.can you find out the value of $a$ & $b$

Can you now finish the problem ……….

From two equations after sim[lificatiopn we get…$a=x$ and $b=\frac{1}{y}$ (as $ab \neq 1)$

Can you finish the problem……..

$\frac{1+b+ab}{a}$=$\frac { 1+\frac{1}{y} +\frac{x}{y}}{x}$=$\frac{(x+y)+1}{xy}=240$

Categories

## Can we Prove that ……..

The length of any side of a triangle is not more than half of its perimeter

### Key Concepts

Triangle Inequality

Perimeter

Geometry

Answer: Yes we can definitely prove that by Triangle Inequality

Mathematical Circles – Chapter 6 – Inequalities Problem 3

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

## Try with Hints

We can start this sum by using this picture below

The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem

b + c > a

So can you try to do the rest of the sum ????????

According to the question we have to find the perimeter at first

Perimeter is the sum of the length of all sides of the triangle = a + b + c

And the length of each side is a or b or c.

We have to prove : a + b + c > length of any one side

This can be one of the most important hint for this problem. Try to do the rest of the sum …………………………..

Here is the rest of the sum :

As stated above if we use triangle inequality :

b + c > a

Lets add a to both the sides

a + b + c > a + a

a + b + c > 2 a

The left hand side of the above inequality is the perimeter of this triangle.

perimeter > 2 a

So , $\frac {perimeter}{2} > a$

$\frac {perimeter}{2}$ = semi perimeter

Hence this is proved that the length of one side of a triangle is less than half of its perimeter.