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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 307.

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

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AIME I Algebra Combinatorics Math Olympiad Math Olympiad Videos USA Math Olympiad

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.

  • 110
  • 420
  • 430
  • 111

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

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AIME I Complex Numbers Math Olympiad Math Olympiad Videos USA Math Olympiad

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.

  • 101
  • 201
  • 301
  • 697

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

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AIME I Math Olympiad USA Math Olympiad

Geometry Problems in AIME; problems and discussions.

Let’s have a problem discussion of Geometry problems in AIME. (American Invitational Mathematics Competitions). Give them a try.

  1. In \Delta ABC, AB = 3, BC = 4 , and CA = 5. Circle \omega intersects\overline{AB} at E and B, \overline{BC} at B and D, and \overline{AC} at F and G. Given that EF=DF and \displaystyle \dfrac{DG}{EG} = \frac{3}{4} , length \displaystyle DE=\dfrac{a\sqrt{b}}{c} , where a and c are relatively prime positive integers, and b is a positive integer not divisible by the square of any prime. Find a+b+c. (2014 AIME I Problems/Problem 15)
  2. Circle C with radius 2 has diameter \overline{AB}. Circle D is internally tangent to circle C at A. Circle E is internally tangent to circle C, externally tangent to circle D, and tangent to \overline{AB}. The radius of circle D is three times the radius of circle E, and can be written in the form \sqrt{m}-n, where m and n are positive integers. Find m+n. (2014 AIME II Problems/Problem 8)
  3. A rectangle has sides of length a and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length a can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length a parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find a^2. (2014 AIME II Problems/Problem 3)
  4. In \triangle{ABC}, AB=10, \angle{A}=30^\circ , and \angle{C=45^\circ}. Let H, D, and M be points on the line BC such that AH\perp{BC}, \angle{BAD}=\angle{CAD}, and BM=CM. Point N is the midpoint of the segment HM, and point P is on ray AD such that PN\perp{BC}.  Then AP^2=\dfrac{m}{n}, where m and n are relatively prime positive integers. Find m+n. (2014 AIME II Problems/Problem 14)
  5. In triangle RED, measured \angle DRE=75^{\circ} and measured \angle RED=45^{\circ}. abs{RD}=1. Let M be the midpoint of segment \overline{RD}. Point C lies on side \overline{ED} such that \overline{RC} \perp \overline{EM}. Extend segment \overline{DE} through E to point A such that CA=AR. Then AE=\frac{a-\sqrt{b}}{c}, where a and c are relatively prime positive integers, and b is a positive integer. Find a+b+c. (2014 AIME II Problems/Problem 11)
  6. In triangle ABC, AB= \frac{20}{11} AC. The angle bisector of angle A intersects BC at point D, and point M is the midpoint of AD. Let P be the point of the intersection of AC and BM. The ratio of CP to PA can be expressed in the form \dfrac{m}{n}, where m and n are relatively prime positive integers. Find m+n. (2011 AIME II Problems/Problem 4)
  7. The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle. (2011 AIME II Problems/Problem 3)
  8. On square ABCD, point E lies on side AD and point F lies on side BC, so that BE=EF=FD=30. Find the area of the square ABCD. (2011 AIME II Problems/Problem 2)
  9. Point P lies on the diagonal AC of square ABCD with AP > CP. Let O_{1} and O_{2} be the circumcenters of triangles ABP and CDP respectively. Given that AB = 12 and {\angle O_{1}PO_{2}} = 120^{\circ}, then AP = \sqrt{a} + \sqrt{b}, where a and b are positive integers. Find a + b. (2011 AIME II Problems/Problem 13)
  10. Gary purchased a large beverage, but only drank m/n of it, where m and n are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only 2/9 as much beverage. Find m+n. (2011 AIME II Problems/Problem 1)
  11. Let ABCDEF be a regular hexagon. Let G, H, I, J, K, and L be the midpoints of sides AB, BC, CD, DE, EF, and AF, respectively. The segments \overbar{AH} , \overbar{BI}, \overbar{CJ}, \overbar{DK}, \overbar{EL}, and \overbar{FG} bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of ABCDEF be expressed as a fraction \frac {m}{n} where m and n are relatively prime positive integers. Find m + n. (2010 AIME II Problems/Problem 9)
  12. Triangle ABC with right angle at C, \angle BAC < 45^\circ and AB = 4. Point P on \overbar{AB} is chosen such that \angle APC = 2\angle ACP and CP = 1. The ratio \frac{AP}{BP} can be represented in the form p + q\sqrt{r}, where p, q, r are positive integers and r is not divisible by the square of any prime. Find p+q+r. (2010 AIME II Problems/Problem 14)
  13. Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is 8: 7. Find the minimum possible value of their common perimeter. (2010 AIME II Problems/Problem 12)
  14. In triangle{ABC} with AB = 12, BC = 13, and AC = 15, let M be a point on \overline{AC} such that the incircles of triangle{ABM} and triangle{BCM} have equal radii. Let p and q be positive relatively prime integers such that \frac {AM}{CM} = \frac {p}{q}. Find p + q. (2010 AIME I Problems/Problem 15)
  15. Rectangle ABCD and a semicircle with diameter AB are coplanar and have nonoverlapping interiors. Let \mathcal{R} denote the region enclosed by the semicircle and the rectangle. Line ell meets the semicircle, segment AB, and segment CD at distinct points N, U, and T, respectively. Line ell divides region \mathcal{R} into two regions with areas in the ratio 1: 2. Suppose that AU = 84, AN = 126, and UB = 168. Then DA can be represented as m\sqrt {n}, where m and n are positive integers and n is not divisible by the square of any prime. Find m + n. (2010 AIME I Problems/Problem 13)
  16. Let \mathcal{R} be the region consisting of the set of points in the coordinate plane that satisfy both |8 - x| + y \le 10 and 3y - x \ge 15. When \mathcal{R} is revolved around the line whose equation is 3y – x = 15, the volume of the resulting solid is \frac {m\pi}{n\sqrt {p}}, where m, n, and p are positive integers, m and n are relatively prime, and p is not divisible by the square of any prime. Find m + n + p. (2010 AIME I Problems/Problem 11)