Squares and Square roots | HANOI 2018

Try this beautiful problem from American Invitational Mathematics Examination, HANOI, 2018 based on Squares and square roots.

Squares and square roots - HANOI 2018

Let a=$(\sqrt2+\sqrt3+\sqrt6)(\sqrt2+\sqrt3-\sqrt6)(\sqrt3+\sqrt6-\sqrt2)(\sqrt6+\sqrt2-\sqrt3)$

b=$(\sqrt2+\sqrt3+\sqrt5)(\sqrt2+\sqrt3-\sqrt5)(\sqrt3+\sqrt5-\sqrt2)(\sqrt5+\sqrt2-\sqrt3)$. The difference a-b belongs to the set

is [-4,0)
is {6}
is [-8,-6]
cannot be determined from the given information

Key Concepts

Algebra

Squares and square roots

Number Theory

Check the Answer

Answer: is [-4,0).

HANOI, 2018

Elementary Number Theory by David Burton

Try with Hints

(x+y+z)(x+y-z)(x-y+z)(-x+y+z)=2$(x^{2}y{2}+y^{2}z^{2}+z^{2}x^{2})-x^{4}-y^{4}-z^{4}$.

We get a-b=2(2+3)(6-5)-$6^{2}+5^{2}$=-1.

Then a-b belongs to [-4,0).

Squares and Inequality | HANOI 2018

Try this beautiful problem from American Invitational Mathematics Examination, HANOI, 2018 based on Squares and inequality.

Squares and inequality - HANOI 2018

Write down all real numbers (x,y) satisfying two conditions $x^{2018}+y^{2}=2$ and $x^{2}+y^{2018}=2$.

is [-1,0)
is (0,1),(-1,0)
is (-1,-1),(-1,1),(1,-1),(1,1)
cannot be determined from the given information

Key Concepts

Algebra

Squares and square roots

Inequality

Check the Answer

Answer: is (-1,-1),(-1,1),(1,-1),(1,1).

HANOI, 2018

Inequalities (Little Mathematical Library) by Korovkin

Try with Hints

If $x^{2}>1$ then$x^{2018}>x^{2}>1$ and $y^{2}<1$ implies that $y^{2} \gt y^{2018}$ Then $x^{2018}+y^{2} \gt x^{2}+y^{2018}$ (contradiction) .

Analogically, if $x^{2} \lt 1$ implies that $x^{2018}+y^{2} \lt x^{2}+y^{2018}$(contradiction).

Then $x^{2}=y^{2}=1$.

Largest Hexagon in Equilateral Triangle | HANOI 2018

Try this beautiful problem from HANOI, 2018 based on Largest Hexagon in Equilateral Triangle.

Geometry - HANOI 2018

Find the largest area of a regular hexagon that can be drawn inside the equilateral triangle of side 3.

is $3\sqrt7$
is $(3\sqrt3)/2$
is $2\sqrt5$
cannot be determined from the given information

Key Concepts

Geometry

Theory of Equations

Number Theory

Check the Answer

Answer: is $(3\sqrt3)/2$.

HANOI, 2018

Geometry Vol I to IV by Hall and Stevens

Try with Hints

Here suppose that the regular hexagon H with side a is inside the triangle equilateral triangle with side 3. Then, the inscribed circle of H is also inside the triangle, and its radius is equal to $(a\sqrt3)/2$

On the other hand, the largest circle in the given equilateral triangle is its inscribed circle whose radius is $(\sqrt3/2)$.

It follows that $a \leq 1$ and the answer is $(6\sqrt3)/4$=$(3\sqrt3)/2$.

Divisibility Problem | HANOI 2018

Try this beautiful problem from American Invitational Mathematics Examination, HANOI, 2018 based on Divisibility.

Divisibility - HANOI 2018

Find all positive integers k such that there exists a positive integer n, for which $2^{n}+11$ is divisible by $2^{k}-1$.

is 2, 4
is 2, 0
is 1,2,4
cannot be determined from the given information

Key Concepts

Number Theory

Inequality

Algebra

Check the Answer

Answer: is 1,2,4.

HANOI, 2018

Elementary Number Theory by David Burton

Try with Hints

Suppose n is a positive integer such that $2^{n}+11 $ is divisible by $2^{k}-1$. Let n=qk+r where q,r are nonnegetive integers and $ 0 \leq r \lt k$. We have $2^{n}+11=2^{kq+r}+11=2^{r}(2^{kq}-1)+(2^{r}+11)$ in comparision with $ 2^{k}-1$ implies $2^{r}+11 $ in comparision with $2^{k}-1$.

Thus we have $2^{r}+11 \geq 2^{k}-1$ and $r \lt k$ Then $k \leq 4$.

By calculating for k=1,2,3,4 we obtain k=1,2,4..

Circles and points | HANOI 2018

Try this beautiful problem from HANOI, 2018 based on Circles and Points.

Circles - HANOI 2018

The center of a circle and nine randomly selected points on this circle are colored in red. Every pair of those points is connected by a line segment, and every point of intersection of two line segments inside the circle is colored in red. Find the largest possible number of red points.

is 224
is 220
is 228
cannot be determined from the given information

Key Concepts

Geometry

Circles

Combination

Check the Answer

Answer: is 220.

HANOI, 2018

Geometry Revisited by Coxeter

Try with Hints

Remark that a convex quadrilateral has exactly one intersection which is the intersection of its two diagonals. Consider 9 points on the circle, which give at most $\frac{9!}{4!5!}$=126 intersections.

Considering the center and three points on the circle, there are at most $\frac{9!}{3!6!}$=84 intersections.

So there are at most 126+84+10=220 red points.

Sequence and Series | HANOI 2018

Try this beautiful problem from HANOI 2018 based on Sequence and Series.

Sequence and Series - HANOI 2018

Let {$u_n$} $n\geq1$ be given sequence satisfying the conditions $u_1=0$, $u_2=1$, $u_{n+1}=u_{n-1}+2n-1$ for $n\geq2$. find $u_{100}+u_{101}$

is 13000
is 10000
is 840
cannot be determined from the given information

Key Concepts

Sequence

Series

Number Theory

Check the Answer

Answer: is 10000.

HANOI, 2018

Principles of Mathematical Analysis by Rudin

Try with Hints

Here $u_2=1$, $u_3=3$, $u_4=6$, $u_5=10$

by induction $u_n=\frac{n(n-1)}{2}$ for every $n\geq1$

Then $u_n+u_{n+1}$=$\frac{n(n-1)}{2}$+$\frac{n(n+1)}{2}$=$n^{2}$ for every $n\geq1$ Then $u_{100}+u_{101}$=10000.

Problem based on Inequalities | HANOI 2018

Try this beautiful problem from HANOI 2018 based on Inequalities.

Inequalities - HANOI 2018

Let a,b,c denote the real numbers such that $1\leq a,b,c \leq 2$. Consider T= $(a-b)^{2018}+(b-c)^{2018}+(c-a)^{2018}$. Find the largest possible value of T.

is 1
is 2
is 8
cannot be determined from the given information

Key Concepts

Inequalities

Algebra

Number Theory

Check the Answer

Answer: is 2.

HANOI, 2018

Inequalities (Little Mathematical Library) by Korovkin

Try with Hints

Here without loss of generality, one can assume that $1 \leq c \leq b \leq a \leq2$ Then $0 \leq a-b \leq 1$ and $(a-b)^{2018} \leq a-b $ and the equality holds if a=b or (a,b)=(2,1)

by the same way $0 \leq b-c \leq 1$ then $(b-c)^{2018} \leq b-c$, $0 \leq a-c \leq 1$ then $(c-a)^{2018} \leq a-c$ Then $T=(a-b)^{2018} + (b-c)^{2018}+ (c-a)^{2018}$

$\leq a-b+b-c+a-c=2(a-c) \leq 2$

The equality holds if (a,b,c)=(2,2,1) or (a,b,c)=(2,1,1). Then max T = 2

Complex Numbers | AIME I, 2009 | Problem 2

Try this beautiful problem from AIME, 2009 based on complex numbers.

Complex Numbers - AIME, 2009

There is a complex number z with imaginary part 164 and a positive integer n such that $\frac{z}{z+n}=4i$, Find n.

Key Concepts

Complex Numbers

Theory of equations

Polynomials

Check the Answer

Answer: 697.

AIME, 2009, Problem 2

Complex Numbers from A to Z by Titu Andreescue .

Try with Hints

Taking z=a+bi

then a+bi=(z+n)4i=-4b+4i(a+n),gives a=-4b b=4(a+n)=4(n-4b)

then n=$\frac{b}{4}+4b=\frac{164}{4}+4.164=697$

Combinations | AIME I, 2009 |Problem 9

Try this problem from American Invitational Mathematics Examination, AIME, 2019 based on Combinations

Combinations- AIME, 2009

A game show offers a contestant three prizes A B and C each of which is worth a whole number of dollars from $1 to $9999 inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A B and C. As a hint the digits of three prizes are given. On a particular day the digits given were 1,1,1,1,3,3,3. Find the total number of possible guesses for all three prizes consistent with the hint.

Key Concepts

Combinations

Theory of equations

Polynomials

Check the Answer

Answer: 420.

AIME I, 2009, Problem 9

Combinatorics by Brualdi .

Try with Hints

Number of possible ordering of seven digits is$\frac{7!}{4!3!}$=35

these 35 orderings correspond to 35 seven-digit numbers, and the digits of each number can be subdivided to represent a unique combination of guesses for A B and C. Thus, for a given ordering, the number of guesses it represents is the number of ways to subdivide the seven-digit number into three nonempty sequences, each with no more than four digits. These subdivisions have possible lengths 1/2/4,2/2/3,1/3/3, and their permutations. The first subdivision can be ordered in 6 ways and the second and third in three ways each, for a total of 12 possible subdivisions.

then total number of guesses is 35.12=420

FERMAT POINT

ABC is a Triangle and P be a Fermat Point Inside it.draw three equilateral triangle based on the three sides i.e$\triangle ABA'$, $\triangle ACC'$, $\triangle BCB'$ respectively.Join $AB'$,$BC'$ and$CA'$ .Show that $ AB'$,$BC'$ and $CA'$ pass through a single piont i.e they are concurrent.

Key Concepts

Rotation

Geometry

shortest distance

Check the Answer

Regional Math Olympiad, India

Challenges and thrills of pre college mathematics

Try with Hints

Rotation:

ABC is a Triangle . Let P Be any point join $AP,BP$ and $CP$. Now if we rotate the $\triangle ABP$ about the point at B $ 60 ^{\circ} $ anti clockwise we will get $\triangle BP'A'$.

SHORTEST DISTANCE:

Join the point P and P'.Now In the triangle BPP' we have

BP-BP'

$\angle PBP'=60 ^{\circ} $, SO $\triangle BPP' $ is a equilateral triangle. so $BP=BP'=PP'$

and also $AP'=AP$ (Length remain unchange after Rotation).

So from the point $A'$ to $C$ the path is $A'P'+PP'+PC$.This path will be Shortest distance if $A'P'+PP'+PC$ i.e A'C be a straight line. and also $AP+PB+AB=A'P'+PP'+PC$

the shortest path betwween two points is a straight line and so $ PA+PB+PC$ reaches its minimum if and only if the point $p$ and $P'$ lie on the line $A'C$

By symmetry it follows that $ P $ must also lie on the line $BC'$ and $AB'$.

So the point of intersection of these lines is a fermat point of a $\triangle ABC$.

EQUILATERAL TRIANGLE :

Now the triangle $AA'B$ we have

$A'B=AB$ (length remain unchange due to rotation)

$\angle A'BA =60^{\circ}$. so the triangle $AA'B $ is a equilateral triangle .

similarly for the other two triangles $AC'C$ and $BB'C$