Categories

Circles and Triangles | AIME I, 2012 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Circles and triangles.

Circles and triangles – AIME I, 2012

Three concentric circles have radii 3,4 and 5. An equilateral triangle with one vertex on each circle has side length s. The largest possible area of the triangle can be written as $a+\frac{b}{c}d^\frac{1}{2}$ where a,b,c,d are positive integers b and c are relative prime and d is not divisible by the square of any prime, find a+b+c+d.

• is 107
• is 41
• is 840
• cannot be determined from the given information

Key Concepts

Angles

Trigonometry

Triangles

AIME I, 2012, Question 13

Geometry Revisited by Coxeter

Try with Hints

In triangle ABC AO=3, BO=4, CO=5 let AB-BC=CA=s [ABC]=$\frac{s^{2}3^\frac{1}{2}}{4}$

$s^{2}=3^{2}+4^{2}-2(3)(4)cosAOB$=25-24cosAOB then [ABC]=$\frac{25(3)^\frac{1}{2}}{4}-6(3)^\frac{1}{2}cosAOB$

of the required form for angle AOB=150 (in degrees) then [ABC]=$\frac{25(3)^\frac{1}{2}}{4}+9$ then a+b+c+d=25+3+4+9=41.

Categories

Arrangement Problem | AIME I, 2012 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Arrangement.

Arrangement – AIME 2012

Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.

• is 107
• is 216
• is 840
• cannot be determined from the given information

Key Concepts

Arrangements

Algebra

Number Theory

AIME, 2012, Question 3

Combinatorics by Brualdi

Try with Hints

Here the number of ways to order the string BBBCCCFFF, such that one B is in first three positions

one C is in 4th to 6th positions, and one F is for last three positions. There are (3)(3)(3)=27 ways for first 3. Then for next two, 2 ways.

Then $(3.2)^3={216}$ways

Categories

Theory of Equations | AIME I, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Theory of Equations.

Theory of Equations – AIME 2015

Let (f(x)) be a third-degree polynomial with real coefficients satisfying[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.]Find (|f(0)|).

• is 107
• is 72
• is 840
• cannot be determined from the given information

Key Concepts

Series

Theory of Equations

Algebra

AIME, 2015, Question 10.

Polynomials by Barbeau.

Try with Hints

Let (f(x)) = (ax^3+bx^2+cx+d). Since (f(x)) is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing (12) and (-12), it is easy to see that (f(1)=f(5)=f(6)), and (f(2)=f(3)=f(7)); otherwise more bends would be required in the graph.

Since only the absolute value of f(0) is required, there is no loss of generalization by stating that (f(1)=12), and (f(2)=-12). This provides the following system of equations.[a + b + c + d = 12][8a + 4b + 2c + d = -12][27a + 9b + 3c + d = -12][125a + 25b + 5c + d = 12][216a + 36b + 6c + d = 12][343a + 49b + 7c + d = -12]

Using any four of these functions as a system of equations yields (|f(0)| = \boxed{072})

Categories

Trigonometry Problem | AIME I, 2015 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Trigonometry.

Trigonometry Problem – AIME 2015

With all angles measured in degrees, the product (\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n), where (m) and (n) are integers greater than 1. Find (m+n).

• is 107
• is 91
• is 840
• cannot be determined from the given information

Key Concepts

Trigonometry

Sequence

Algebra

AIME, 2015, Question 13.

Plane Trigonometry by Loney .

Try with Hints

Let (x = \cos 1^\circ + i \sin 1^\circ). Then from the identity[\sin 1 = \frac{x – \frac{1}{x}}{2i} = \frac{x^2 – 1}{2 i x},]we deduce that (taking absolute values and noticing (|x| = 1))[|2\sin 1| = |x^2 – 1|.]

But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ – z)$, if we let our product be $M$ then[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ][= \frac{1}{2^{90}} |x^2 – 1| |x^6 – 1| |x^{10} – 1| \dots |x^{354} – 1| |x^{358} – 1|]because $\sin$ is positive in the first and second quadrants.

Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z – x^2)(z – x^6)\dots (z – x^{358}) = z^{90} + 1$, and so[\frac{1}{M} = \dfrac{1}{2^{90}}|1 – x^2| |1 – x^6| \dots |1 – x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.]It is easy to see that $M = 2^{89}$ and that our answer is $2+89=91$.

Categories

Smallest Perimeter of Triangle | AIME I, 2015 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Smallest Perimeter of Triangle.

Smallest Perimeter of Triangle – AIME 2015

Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\triangle ABC$..

• is 107
• is 108
• is 840
• cannot be determined from the given information

Key Concepts

Inequalities

Trigonometry

Geometry

AIME, 2015, Question 11

Geometry Vol I to IV by Hall and Stevens

Try with Hints

Let $D$ be the midpoint of $\overline{BC}$. Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$, so $\angle ADB = \angle ADC = 90^o$.Now let $BD=y$, $AB=x$, and $\angle IBD$ =$\frac{\angle ABD}{2}$ = $\theta$.Then $\mathrm{cos}{(\theta)} = \frac{y}{8}$and $\mathrm{cos}{(2\theta)} = \frac{y}{x} = 2\mathrm{cos^2}{(\theta)} – 1 = \frac{y^2-32}{32}$.

Cross-multiplying yields $32y = x(y^2-32)$.

Since $x,y>0$, $y^2-32$ must be positive, so $y > 5.5$.

Additionally, since $\triangle IBD$ has hypotenuse $\overline{IB}$ of length $8$, $BD=y < 8$.

Therefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$, $6.5$, $7$, and $7.5$.

However, only one of these values, $y=6$, yields an integral value for $AB=x$, so we conclude that $y=6$ and $x=\frac{32(6)}{(6)^2-32}=48$.

Thus the perimeter of $\triangle ABC$ must be $2(x+y) = {108}$.

Categories

Cube of Positive Integer | Number Theory | AIME I, 2015 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Cube of Positive Integer.

Cube of Positive Numbers – AIME I, 2015

There is a prime number p such that 12p+1 is the cube of positive integer.Find p..

• is 107
• is 183
• is 840
• cannot be determined from the given information

Key Concepts

Algebra

Theory of Equations

Number Theory

AIME, 2015, Question 3

Elementary Number Theory by David Burton

Try with Hints

$a^{3}=12p+1$ implies that $a^{3}-1=12p$ that is (a-1)($a^{2}$+a+1)=12p

a is odd, a-1 even, $a^{2} +a+1 odd implies a-1 multiple of 12 that is here =12 then a=12+1 =13 \(a^{2}+a+1=p implies p= 169+13+1=183. Subscribe to Cheenta at Youtube Categories Number and Series | Number Theory | AIME I, 2015 Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Number Theory and Series. Number Theory and Series – AIME 2015 The expressions A = \((1 \times 2)+(3 \times 4)+….+(35 \times 36)+37$ and B = $1+(2 \times 3)+(4 \times 5)+….+(36 \times 37)$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers A and B.

• is 107
• is 648
• is 840
• cannot be determined from the given information

Key Concepts

Series

Theory of Equations

Number Theory

AIME, 2015, Question 1

Elementary Number Theory by David Burton

Try with Hints

B-A=$-36+(2 \times 3)+….+(2 \times 36)$

=$-36+4 \times (1+2+3+….+18)$

=$-36+(4 \times \frac{18 \times 19}{2})$=648.

Categories

Probability Biased and Unbiased | AIME I, 2010 Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2010 based on Probability Biased and Unbiased.

Probability Biased and Unbiased – AIME 2010

Ramesh and Suresh have two fair coins and a third coin that comes up heads with probability $\frac{4}{7}$,Ramesh flips the three coins, and then Suresh flips the three coins, let $\frac{m}{n}$ be the probability that Ramesh gets the same number of heads as Suresh, where m and n are relatively prime positive integers. Find m+n.

• is 107
• is 250
• is 840
• cannot be determined from the given information

Key Concepts

Series

Probability

Number Theory

AIME, 2010, Question 4

Combinatorics by Brualdi

Try with Hints

No heads TTT is $\frac{1.1.1}{2.2.7}=\frac{3}{28}$and $(\frac{3}{28})^{2}=\frac{9}{784}$

One Head HTT THT TTH with $\frac{3}{28}$ $\frac {3}{28}$ and $\frac{4}{28}$ then probability is $\frac{4(3.3)+4(3.4)+1(4.4)}{28^{2}}$=$\frac{100}{784}$

Two heads HHT $\frac{4}{28}$ HTH $\frac{4}{28}$ THH $\frac{3}{28}$ then probability is $\frac{1(3.3)+4(3.4)+4(4.4)}{28^{2}}$=$\frac{121}{784}$.

Three heads HHH is $\frac{4}{28}$ then probability $\frac{16}{784}$

Then sum is $\frac{9+100+121+16}{784}=\frac{123}{392}$ then 123+392=515.

Categories

Number Theory of Primes | AIME I, 2015

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Number Theory of Primes.

Number Theory of Primes – AIME 2015

There is a prime number p such that 16p+1 is the cube of a positive integer. Find p.

• is 307
• is 250
• is 840
• cannot be determined from the given information

Key Concepts

Series

Theory of Equations

Number Theory

AIME, 2015

Elementary Number Theory by Sierpinsky

Try with Hints

Notice that 16p+1must be in the form $(a+1)^{3}=a^{3}+3a^{2}+3a$, or $16p=a(a^{2}+3a+3)$. Since p must be prime, we either have p=a or a=16

p not equal to a then we have a=16,

p\(=16^{2}+3(16)+3=307