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Geometry Math Olympiad PRMO USA Math Olympiad

Area of the Trapezium | PRMO-2017 | Question 30

Try this beautiful Problem from Geometry based on Area of the Trapezium from PRMO 2017.

Area of the Trapezium – PRMO 2017, Problem 30


Consider the areas of the four triangles obtained by drawing the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a trapezium ABCD. The product of these areas, taken two at time, are computed. If among the six products so obtained, two products are 1296 and 576 , determine the square root of the maximum possible area of the trapezium to the nearest integer.

  • $9$
  • $40$
  • $13$
  • $20$

Key Concepts


Geometry

Triangle

Trapezium

Check the Answer


Answer:$13$

PRMO-2017, Problem 30

Pre College Mathematics

Try with Hints


Let $x, y, z, w$ be areas of the four triangles as shown in figure.
then area of \(\triangle ADB\)= Area of \(\triangle ACB\)
$\Rightarrow x+y=x+w \Rightarrow y=w$

Also \(\frac{AE}{EC}\)=\(\frac{area of \triangle ADE}{area of \triangle DEF}\)=\(\frac{area of \triangle AEB}{area of \triangle BEC}\)
$\Rightarrow \frac{y}{z}=\frac{x}{w}=\frac{x}{y} \Rightarrow y^{2}=z x$
$\Rightarrow z, y, x$ are in G.P.

Can you now finish the problem ……….

Let $y=z r$ and $x=z r^{2},$ where $r \geq 1$ To make area of trapezium ABCD maximum, we take $z y=z^{2} r=576$
and $y w=z^{2} r^{2}=1296$
As $( z \leq y \leq x)$
Therefore $ \frac{z^{2} r^{2}}{z^{2} r}=\frac{1296}{576} \Rightarrow r=\frac{9}{4} \Rightarrow z=16$

Can you finish the problem……..

Therefore area of trapezium $\mathrm{ABCD}$
$=x+y+z+w=z r^{2}+2 z r+z$
$=z(1+r)^{2}=16\left(1+\frac{9}{4}\right)^{2}=13^{2}$
Therefore Answer is $13 .$

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Categories
Algebra Math Olympiad PRMO USA Math Olympiad

Positive Integer | PRMO-2017 | Question 1

Try this beautiful Positive Integer Problem from Algebra from PRMO 2017, Question 1.

Positive Integer – PRMO 2017, Question 1


How many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by $3 ?$

  • $9$
  • $7$
  • $28$

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:$28$

PRMO-2017, Problem 1

Pre College Mathematics

Try with Hints


Let $n$ be the positive integer less than 1000 and $s$ be the sum of its digits, then $3 \mid n$ and $7 \mid s$
$3|n \Rightarrow 3| s$
therefore$21| s$

Can you now finish the problem ……….

Also $n<1000 \Rightarrow s \leq 27$
therefore $\mathrm{s}=21$
Clearly, n must be a 3 digit number Let $x_{1}, x_{2}, x_{3}$ be the digits, then $x_{1}+x_{2}+x_{3}=21$
where $1 \leq x_{1} \leq 9,0 \leq x_{2}, x_{3} \leq 9$
$\Rightarrow x_{2}+x_{3}=21-x_{1} \leq 18$
$\Rightarrow x_{1} \geq 3$

Can you finish the problem……..

For $x_{1}=3,4, \ldots ., 9,$ the equation (1) has $1,2,3, \ldots ., 7$ solutions
therefore total possible solution of equation (1)

=$1+2+\ldots+7=\frac{7 \times 8}{2}=28$

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Algebra Math Olympiad PRMO USA Math Olympiad

Rectangle Problem | Geometry | PRMO-2017 | Question 13

Try this beautiful Rectangle Problem from Geometry, from PRMO 2017.

Rectangle Problem – Geometry – PRMO 2017, Question 13


In a rectangle $A B C D, E$ is the midpoint of $A B ; F$ is a point on $A C$ such that $B F$ is perpendicular to $A C$; and FE perpendicular to BD. Suppose $\mathrm{BC}=8 \sqrt{3}$. Find AB.

  • $9$
  • $24$
  • $11$

Key Concepts


Geometry

Triangle

Trigonometry

Check the Answer


Answer:$24$

PRMO-2017, Problem 13

Pre College Mathematics

Try with Hints


Rectangle Problem

We have to find out the value of \(AB\). Join \(BD\). \(BF\) is perpendicular on \(AC\).

Let $\angle \mathrm{BAC}=\theta$
since $\mathrm{E}$ is mid point of hypotenous $\mathrm{AB}$ of right $\Delta \mathrm{AFB}$, therefore $A E=F E=B E$

Can you now finish the problem ……….

Rectangle Problem

Therefore

Therefore$ \angle E F A=\angle F A E=\theta$
and $\angle \mathrm{FEB}=\angle \mathrm{EAF}+\angle \mathrm{EFA}=2 \theta$
$\Rightarrow \angle E B D=90^{\circ}-\angle B E F=90^{\circ}-2 \theta$
But $\angle \mathrm{FAE}=\angle \mathrm{CAB}=\angle \mathrm{DBA}$
Therefore $\theta=90^{\circ}-2 \theta \Rightarrow \theta=30^{\circ}$
Therefore in $\Delta \mathrm{ABC}, \tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}$

Can you finish the problem……..

Therefore $A B=B C \cot \theta=8 \sqrt{3} \cot 30^{\circ}=24$

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Categories
Algebra Math Olympiad PRMO USA Math Olympiad

Pen & Note Books Problem| PRMO-2017 | Question 8

Try this beautiful Pen & Note Books Problem from Algebra, from PRMO 2017.

Pen & Note Books – PRMO 2017, Question 8


A pen costs $Rs 11$ and a notebook costs $Rs. 13 .$ Find the number of ways in which a person can spend exactly Rs.1000 to buy pens and notebooks.

  • $9$
  • $7$
  • $11$

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:$7$

PRMO-2017, Problem 8

Pre College Mathematics

Try with Hints


Given A pen costs Rs.\(11\) and a note book costs Rs.\(13\)

$11 x+13 y=1000$………………….(1)

Can you now finish the problem ……….

Now $11 x+13 y=1000$
$\Rightarrow 11 x=1000-13 y=(1001-11 y)-(2 y+1)$

=$11(91-y)-(2 y+1)$
$\Rightarrow 11 | 2 y+1$
Let $2 y+1=11(2 k-1), k \in I^{+}$
$\Rightarrow y=11 \mathrm{k}-6$
therefore $11 x=11(97-11 k)-11(2 k-1)$
$\Rightarrow x=98-13 k$
But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$

Can you finish the problem……..

But $x>0 \Rightarrow k<\frac{98}{13} \Rightarrow k \leq 7$
therefore for each $\mathrm{k} \in{1,2, \ldots ., 7},$ we get a unique pair $(\mathrm{x}, \mathrm{y})=(98-13 \mathrm{k}, 11 \mathrm{k}-6)$ satisfying equation
Hence 7 ways are possible.

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Categories
Algebra Math Olympiad PRMO USA Math Olympiad

Chords in a Circle | PRMO-2017 | Question 26

Try this beautiful Problem based on Chords in a Circle, Geometry, from PRMO 2017.

Chords in a Circle – PRMO 2017, Question 26


Let $A B$ and $C D$ be two parallel chords in a circle with radius 5 such that the centre $O$ lies between these chords. Suppose $A B=6, C D=8 .$ Suppose further that the area of the part of the circle lying between the chords $A B$ and $C D$ is $(m \pi+n) / k,$ where $m, n, k$ are positive integers with gcd$(m, n, k)=1$ . What is the value of $m+n+k ?$

  • $9$
  • $75$
  • $11$

Key Concepts


Geometry

Triangle

Circle

Check the Answer


Answer:$75$

PRMO-2017, Problem 26

Pre College Mathematics

Try with Hints


Chords in a Circle

Draw OE $\perp A B$ and $O F \perp C D$

Clearly $\mathrm{EB}=\frac{\mathrm{AB}}{2}=3, \mathrm{FD}=\frac{\mathrm{CU}}{2}=4$

$\mathrm{OE}=\sqrt{5^{2}-3^{2}}=4$ and $\mathrm{OF}=\sqrt{5^{2}-4^{2}}=3$

Therefore $\Delta O E B \sim \Delta D F O$

Can you now finish the problem ……….

Chords in a Circle

Let $\angle \mathrm{EOB}=\angle \mathrm{ODF}=\theta,$ then

$\angle B O D=\angle A O C=180^{\circ}-\left(\theta+90^{\circ}-\theta\right)=90^{\circ}$

Now area of portion between the chords

= \(2 \times\) (area of minor sector BOD)+2 \times ar\((\triangle AOB)\)
$=2 \times \frac{\pi \times 5^{2}}{4}+2 \times \frac{1}{2} \times 6 \times 4=\frac{25 \pi}{2}+24=\frac{25 \pi+48}{2}$

Therefore $m=25, n=48$ and $k=2$

Can you finish the problem……..

Therefore $m+n+k=75$

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Algebra Math Olympiad PRMO USA Math Olympiad

Circle | Geometry Problem | PRMO-2017 | Question 27

Try this beautiful Problem from Geometry based on Circle from PRMO 2017.

Circle – PRMO 2017, Problem 27


Let $\Omega_{1}$ be a circle with centre 0 and let $A B$ be a diameter of $\Omega_{1} .$ Le $P$ be a point on the segment $O B$ different from 0. Suppose another circle $\Omega_{2}$ with centre P lies in the interior of $\Omega_{1}$. Tangents are drawn from A and B to the circle $\Omega_{2}$ intersecting $\Omega_{1}$ again at $A_{1}$ and $B_{1}$ respectively such that $A_{1}$, and $B_{1}$ are on the opposite sides of $A B$. Given that $A_{1} B=5, A B_{1}=15$ and $O P=10,$ find the radius of $\Omega_{1}$

  • $9$
  • $40$
  • $34$
  • $20$

Key Concepts


Geometry

Circle

Check the Answer


Answer:$20$

PRMO-2017, Problem 27

Pre College Mathematics

Try with Hints


Circle Problem

Let radius of $\Omega_{1}$ be $R$ and that of $\Omega_{2}$ be $r$
From figure, $\Delta \mathrm{ADP} \sim \Delta \mathrm{AA}_{1} \mathrm{B}$
[
\begin{array}{l}
\Rightarrow \frac{D P}{A, B}=\frac{A P}{A B} \
\Rightarrow \frac{r}{5}=\frac{R+10}{2 R}
\end{array}
]

Can you now finish the problem ……….

Circle Problem figure

Again, $\Delta B P E \sim \Delta B A B_{1}$
Therefore $\frac{P E}{A B_{1}}=\frac{B P}{B A}$
$\Rightarrow \frac{r}{15}=\frac{R-10}{2 R}$

Can you finish the problem……..

Dividing (1) by (2)

$3=\frac{R+10}{R-10} \Rightarrow R=20$

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Categories
Algebra Math Olympiad PRMO USA Math Olympiad

Pen & Note Books Problem | PRMO-2019 | Question 16

Try this beautiful Problem from Algebra based on Pen & Note Books from PRMO 2019.

Pen & Note Books – PRMO 2019, Problem 16


A pen costs Rs.13 and a notebook costs Rs.17. A school spends exactly Rs.10000 in the year 2017 – 18 to buy x pens and y notebooks such that $x$ and $y$ are as close as possible (i.e. $|x-y|$ is minimum). Next year, in $2018-19,$ the school spends a little more than Rs.10000 and buys y pens and x notebooks. How much more did the school pay?

  • $9$
  • $40$
  • $34$

Key Concepts


Algebra

Equation

multiplication

Check the Answer


Answer:$40$

PRMO-2019, Problem 16

Pre College Mathematics

Try with Hints


Given A pen costs Rs.\(13\) and a note book costs Rs.\(17\)

Nunber of pens \(x\) and numbers of notebooks \(y\)

According to question

\(13 x+17 y=10,000\) ……………………..(1)
\(17 x+13 y=10000+P\)………………….(2)

Where $\mathrm{P}$ is little more amount spend by school in $2018-19$.we have to find out the value of \(P\).Can you find out?

Can you now finish the problem ……….

Adding (1) & (2)

$x+y=\frac{20,000+P}{30}$

Subtract (1) form (2)

$x-y=\frac{P}{4}$

Let $P=4 a$
$\begin{aligned} \text { So, } x+y=& \frac{20,000+4 a}{30} \ &=666+\frac{10+2 a}{15} \end{aligned}$

Therefore \(a\)=\(10\)

Can you finish the problem……..

Since $P=4 a$ \(\Rightarrow P=40\)

So school had to pay Rs. 40 extra in 2018 – 2019

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Categories
India Math Olympiad Math Olympiad Number Theory PRMO USA Math Olympiad

Sum of digits Problem | PRMO 2016 | Question 6

Try this beautiful problem from Number system based on sum of digits.

Sum of digits | PRMO | Problem 6


Find the sum of digits in decimal form of the number \((9999….9)^3\) (There are 12 nines)

  • $200$
  • $216$
  • $230$

Key Concepts


Number system

Digits

counting

Check the Answer


Answer:$216$

PRMO-2016, Problem 6

Pre College Mathematics

Try with Hints


we don’t know what will be the expression of \((9999….9)^3\). so we observe….

\(9^3\)=\(729\)

\((99)^3\)=\(970299\)

\((999)^3\)=\(997002999\)

……………..

……………

we observe that,There is a pattern such that…

In \((99)^3\)=\(970299\) there are 1-nine,1-seven,1-zero,1-two,2-nines & \((999)^3\)=\(997002999\) there are 2- nines,1-seven,2-zeros,1-two,3-nines….so in this way…..\((999….9)^3\) will be 11-nines,1-seven,11-zeros,1-two,12-nines……….

Therefore \((999….9)^3\)=\((99999999999) 7 (00000000000) 2(999999999999)\)

can you finish the problem?

Therefore \((999….9)^3\)=\((99999999999) 7 (00000000000) 2(999999999999)\)…….

total numbers of Nines are (11+12) and (7+2)=9(another one) …..so total (11+12+1)=24 nines and the sum be \((24\times 9)\)=\(216\)

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Categories
Math Olympiad Number Theory PRMO USA Math Olympiad

Integer based Problem | PRMO-2018 | Question 20

Try this beautiful Integer based Problem from Algebra, from PRMO 2018.

Integer based Problem – PRMO 2018, Question 20


Determine the sum of all possible positive integers n, the product of whose digits equals \(n^2 -15n – 27 \)

  • $9$
  • $17$
  • $34$

Key Concepts


Algebra

Integer

multiplication

Check the Answer


Answer:$17$

PRMO-2018, Problem 17

Pre College Mathematics

Try with Hints


Product of digits = \(n^2 – 15n – 27 = n(n – 15) – 27\)

so at first we observe when n=one digit ,two digit and 3 digit numbers…..

If n is a more than 2-digit number, say 3-digit number, then product has to be\(\leq 9 × 9 × 9 = 729\) but \((n(n – 15) – 27)\) is more than 729 (in fact it a more than 3-digit numbers for any 3-digit n). Hence, n can be either one-digit or 2-digit.

If n is 1-digit then \(n^2 – 15n – 27 = n\) \(\Rightarrow n\)= not an integer , so n is a two digit number

now we will observe for 2-digit numbers…..

Can you now finish the problem ……….

For Two-digit numbers:

As product is positive so n(n-15)-27>0\(\Rightarrow n\geq 17\)

Now two digit product is less than equal to 81

so \(n(n-15)-27\leq 1\)\(\Rightarrow n(n-15)\leq 108\) \(\Rightarrow n\leq 20\)

Therefore n can be \(17\),\(18\),\(19\) or \(20\)

Can you finish the problem……..

For \(n\)= \(17\),\(18\),\(19\) or \(20\)

when n=17,then \(n(n-15)-27=7=1 \times 7\)

when n=18,then \(n(n-15)-27=27\neq 1\times 8\)

when n=19,then \(n(n-15)-27=49=1 \neq 9\)

when n=20,then \(n(n-15)-27=73=1 \neq 0\)

Therefore \(n\)=17

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India Math Olympiad Math Olympiad PRMO USA Math Olympiad

Sum of two digit numbers | PRMO-2016 | Problem 7

Try this beautiful problem from Algebra based on Sum of two digit numbers from PRMO 2016.

Sum of two digit numbers | PRMO | Problem 7


Let s(n) and p(n) denote the sum of all digits of n and the products of all the digits of n(when written in decimal form),respectively.Find the sum of all two digits natural numbers n such that \(n=s(n)+p(n)\)

  • $560$
  • $531$
  • $654$

Key Concepts


Algebra

number system

addition

Check the Answer


Answer:$531$

PRMO-2016, Problem 7

Pre College Mathematics

Try with Hints


Let \(n\) is a number of two digits ,ten’s place \(x\) and unit place is \(y\).so \(n=10x +y\).given that \(s(n)\)= sum of all digits \(\Rightarrow s(n)=x+y\) and \(p(n)\)=product of all digits=\(xy\)

now the given condition is \(n=s(n)+p(n)\)

Can you now finish the problem ……….

From \(n=s(n)+p(n)\) condition we have,

\(n=s(n)+p(n)\) \(\Rightarrow 10x+y=x+y+xy \Rightarrow 9x=xy \Rightarrow y=9\) and the value of\(x\) be any digit….

Can you finish the problem……..

Therefore all two digits numbers are \(19,29,39,49,59,69,79,89,99\) and sum=\(19+29+39+49+59+69+79+89+99=531\)

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