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Triangle and Trigonometry | AIME I, 1999 Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Triangle and Trigonometry.

Triangle and Trigonometry – AIME 1999

Point P is located inside triangle ABC so that angles PAB,PBC and PCA are all congruent. The sides of the triangle have lengths AB=13, BC=14, CA=15, and the tangent of angle PAB is $\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 463
• is 840
• cannot be determined from the given information

Key Concepts

Triangles

Angles

Trigonometry

AIME, 1999, Question 14

Geometry Revisited by Coxeter

Try with Hints

Let y be the angleOAB=angleOBC=angleOCA then from three triangles within triangleABC we have $b^{2}=a^{2}+169-26acosy$ $c^{2}=b^{2}+196-28bcosy$ $a^{2}=c^{2}+225-30ccosy$ adding these gives cosy(13a+14b+15c)=295

[ABC]=[AOB]+[BOC]+[COA]=$\frac{siny(13a+14b+15c)}{2}$=84 then (13a+14b+15c)siny=168

tany=$\frac{168}{295}$ then 168+295=463.

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Probability in Games | AIME I, 1999 | Question 13

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 1999 based on Probability in games.

Probability in Games – AIME I, 1999 Question 13

Forty teams play a tournament in which every team plays every other team exactly once. No ties occur,and each team has a 50% chance of winning any game it plays.the probability that no two team win the same number of games is $\frac{m}{n}$ where m and n are relatively prime positive integers, find $log_{2}n$

• 10
• 742
• 30
• 11

Key Concepts

Probability

Theory of equations

Combinations

AIME, 1999 Q13

Course in Probability Theory by Kai Lai Chung .

Try with Hints

${40 \choose 2}$=780 pairings with $2^{780}$ outcomes

no two teams win the same number of games=40! required probability =$\frac{40!}{2^{780}}$

the number of powers of 2 in 40!=[$\frac{40}{2}$]+[$\frac{40}{4}$]+[$\frac{40}{8}$]+[$\frac{40}{16}$]+[$\frac{40}{32}$]=20+10+5+2+1=38 then 780-38=742.

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Least Positive Integer Problem | AIME I, 2000 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Least Positive Integer.

Least Positive Integer Problem – AIME I, 2000

Find the least positive integer n such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit 0.

• is 107
• is 8
• is 840
• cannot be determined from the given information

Key Concepts

Product

Least positive integer

Integers

AIME I, 2000, Question 1

Elementary Number Theory by Sierpinsky

Try with Hints

$10^{n}$ has factor 2 and 5

for n=1 $2^{1}$=2 $5^{1}$=5

for n=2 $2^{2}$=4 $5^{2}=25$

for n=3 $2^{3}$=8 $5^{3}=125$

……..

for n=8 $2^{8}$=256 $5^{8}=390625$

here $5^{8}$ contains the zero then n=8.

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Equations and Complex numbers | AIME I, 2019 Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2019 based on Equations and Complex numbers.

Equations and Complex numbers – AIME 2019

For distinct complex numbers $z_1,z_2,……,z_{673}$ the polynomial $(x-z_1)^{3}(x-z_2)^{3}…..(x-z_{673})^{3}$ can be expressed as $x^{2019}+20x^{2018}+19x^{2017}+g(x)$, where g(x) is a polynomial with complex coefficients and with degree at most 2016. The value of $|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|$ can be expressed in the form $\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n

• is 107
• is 352
• is 840
• cannot be determined from the given information

Key Concepts

Equations

Complex Numbers

Integers

AIME, 2019, Question 10

Complex Numbers from A to Z by Titu Andreescue

Try with Hints

here $|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|$=s=$(z_1z_2+z_1z_3+….z_1z_{673})+(z_2z_3+z_2z_4+…+z_2z_{673})$

$+…..+(z_{672}z_{673})$ here

P=$(x-z_1)(x-z_1)(x-z_1)(x-z_2)(x-z_2)(x-z_2)…(x-z_{673})(x-z_{673})(x-z_{673})$

with Vieta’s formula,$z_1+z_1+z_1+z_2+z_2+z_2+…..+z_{673}+z_{673}+z_{673}$=-20 then $z_1+z_2+…..+z_{673}=\frac{-20}{3}$ the first equation and ${z_1}^{2}+{z_1}^{2}+{z_1}^{2}+{z_1z_2}+{z_1z_2}+{z_1z_2}+…..$=$3({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})$+$9({z_1z_2}+{z_1z_3}+….+{z_{672}z_{673}})$=$3({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})$+9s which is second equation

here $(z_1+z_2+…..+z_{673})^{2}=\frac{400}{9}$ from second equation then $({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})+2({z_1z_2}+{z_1z_3}+….+{z_{672}z_{673}})=\frac{400}{9}$ then $({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})+2s=\frac{400}{9}$ then $({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})=\frac{400}{9}$-2s then with second equation and with vieta s formula $3(\frac{400}{9}-2s)+9s$=19 then s=$\frac{-343}{9}$ then |s|=$\frac{343}{9}$ where 343 and 9 are relatively prime then 343+9=352.

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Probability of divisors | AIME I, 2010 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Probability of divisors.

Probability of divisors – AIME I, 2010

Ramesh lists all the positive divisors of $2010^{2}$, she then randomly selects two distinct divisors from this list. Let p be the probability that exactly one of the selected divisors is a perfect square. The probability p can be expressed in the form $\frac{m}{n}$, 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 I, 2010, Question 1

Elementary Number Theory by Sierpinsky

Try with Hints

$2010^{2}=2^{2}3^{2}5^{2}67^{2}$

$(2+1)^{4}$ divisors, $2^{4}$ are squares

probability is $\frac{2.2^{4}.(3^{4}-2^{4})}{3^{4}(3^{4}-1)}=\frac{26}{81}$ implies m+n=107

Categories

Area of Equilateral Triangle | AIME I, 2015 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 from Geometry, based on Area of Equilateral Triangle (Question 4).

Area of Triangle – AIME I, 2015

Point B lies on line segment AC with AB =16 and BC =4. Points D and E lie on the same side of line AC forming equilateral triangle ABD and traingle BCE. Let M be the midpoint of AE, and N be the midpoint of CD. The area of triangle BMN is x. Find $x^{2}$.

• is 107
• is 507
• is 840
• cannot be determined from the given information

Key Concepts

Algebra

Theory of Equations

Geometry

AIME, 2015, Question 4

Geometry Revisited by Coxeter

Try with Hints

Let A(0,0), B(16,0),C(20,0). let D and E be in first quadrant. then D =$(8,8\sqrt3)$, E=$(18,2\sqrt3$).

M=$(9,\sqrt3)$, N=($14,4\sqrt3$), where M and N are midpoints

since BM, BN, MN are all distance, BM=BN=MN=$2\sqrt13$. Then, by area of equilateral triangle, x=$13\sqrt3$ then$x^{2}$=507.

Categories

Probability Problem | Combinatorics | AIME I, 2015 – Question 5

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

Probability Problem – AIME I, 2015

In a drawer Sandy has 5 pairs of socks, each pair a different color. on monday sandy selects two individual socks at random from the 10 socks in the drawer. On tuesday Sandy selects 2 of the remaining 8 socks at random and on wednesday two of the remaining 6 socks at random. The probability that wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 341
• is 840
• cannot be determined from the given information

Key Concepts

Algebra

Theory of Equations

Probability

AIME, 2015, Question 5

Geometry Revisited by Coxeter

Try with Hints

Wednesday case – with restriction , select the pair on wednesday in $5 \choose 1$ ways

Tuesday case – four pair of socks out of which a pair on tuesday where a pair is not allowed where 4 pairs are left,the number of ways in which this can be done is $8 \choose 2$ – 4

Monday case – a total of 6 socks and a pair not picked $6 \choose 2$ -2

by multiplication and principle of combinatorics $\frac{(5)({5\choose 2} -4)({6 \choose 2}-2)}{{10 \choose 2}{8 \choose 2}{6 \choose 2}}$=$\frac{26}{315}$. That is 341.

Categories

Probability of tossing a coin | AIME I, 2009 | Question 3

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on Probability of tossing a coin.

Probability of tossing a coin – AIME I, 2009 Question 3

A coin that comes up heads with probability p>0and tails with probability (1-p)>0 independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to $\frac{1}{25}$ the probability of five heads and three tails. Let p=$\frac{m}{n}$ where m and n are relatively prime positive integers. Find m+n.

• 10
• 20
• 30
• 11

Key Concepts

Probability

Theory of equations

Polynomials

AIME, 2009

Course in Probability Theory by Kai Lai Chung .

Try with Hints

here $\frac{8!}{3!5!}p^{3}(1-p)^{5}$=$\frac{1}{25}\frac{8!}{5!3!}p^{5}(1-p)^{3}$

then $(1-p)^{2}$=$\frac{1}{25}p^{2}$ then 1-p=$\frac{1}{5}p$

then p=$\frac{5}{6}$ then m+n=11

Categories

Equations with number of variables | AIME I, 2009 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2009 based on Equations with a number of variables.

Equations with number of variables – AIME 2009

For t=1,2,3,4, define $S^{t}=a^{t}_1+a^{t}_2+…+a^{t}_{350}$, where $a_{i}\in${1,2,3,4}. If $S_{1}=513, S_{4}=4745$, find the minimum possible value for $S_{2}$.

• is 905
• is 250
• is 840
• cannot be determined from the given information

Key Concepts

Series

Theory of Equations

Number Theory

AIME, 2009, Question 14

Polynomials by Barbeau

Try with Hints

j=1,2,3,4, let $m_{j}$ number of $a_{i}$ s = j then $m_{1}+m{2}+m{3}+m{4}=350$, $S_{1}=m_{1}+2m_{2}+3m_{3}+4m_{4}=513$ $S_{4}=m_{1}+2^{4}m_{2}+3^{4}m_{3}+4^{4}m_{4}=4745$

Subtracting first from second, then first from third yields $m_{2}+2m_{3}+3m_{4}=163,$ and $15m_{2}+80m_{3}+255m_{4}=4395$ Now subtracting 15 times first from second gives $50m_{3}+210m_{4}=1950$ or $5m_{3}+21m_{4}=195$ Then $m_{4}$ multiple of 5, $m_{4}$ either 0 or 5

If $m_{4}=0$ then $m_{j}$ s (226,85,39,0) and if $m_{4}$=5 then $m_{j}$ s (215,112,18,5) Then $S_{2}=1^{2}(226)+2^{2}(85)+3^{2}(39)+4^{2}(0)=917$ and $S_{2}=1^{2}(215)+2^{2}(112)+3^{2}(18)+4^{2}(5)=905$ Then min 905.

Categories

Arrangement of digits | AIME I, 2012 | Question 5

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

Arrangement of digits – AIME 2012

Let B be the set of all binary integers that can be written using exactly 5 zeros and 8 ones where leading zeros are allowed. If all possible subtractions are performed in which one element of B is subtracted from another, find the number of times the answer 1 is obtained.

• is 107
• is 330
• is 840
• cannot be determined from the given information

Key Concepts

Arrangements

Algebra

Number Theory

AIME, 2012, Question 5

Combinatorics by Brualdi

Try with Hints

When 1 subtracts from a number, the number of digits remain constant when the initial number has units and tens place in 10

Then for subtraction from B requires one number with unit and tens place 10.

10 there, remaining 1 distribute any of other 11 then answer ${11 \choose 7} = {330}$.