Difference between revisions of "User:Ddk001"
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*Note: All the solutions so far have been made by me :) | *Note: All the solutions so far have been made by me :) | ||
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+ | <math>\textbf{I wrote a couple of solutions here. Hope it's okay :) ~ cxsmi (please feel free to delete this note and/or the solutions)}</math> | ||
===Problem 1=== | ===Problem 1=== | ||
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===Solution 2=== | ===Solution 2=== | ||
− | Recall that a perfect square <math>(a + b)^2</math> can be written as <math>a^2 + 2ab + b^2</math>. Since <math>m^2</math> is a perfect square, the <math>RHS</math> must be in this form. We substitute <math>2^4</math> for <math>a</math> to get that <math>2^8 + 2^5 \cdot 2^b + 2^{2b} = m^2</math>. To make the middle term have an exponent of <math>11</math>, we must have <math>b = 6</math>. Then <math>n = 12</math> and <math>m = (2^4 + 2^6)^2 = (16 + 64)^2 = 80^2</math>, so <math>m + n = \boxed{ | + | Recall that a perfect square <math>(a + b)^2</math> can be written as <math>a^2 + 2ab + b^2</math>. Since <math>m^2</math> is a perfect square, the <math>RHS</math> must be in this form. We substitute <math>2^4</math> for <math>a</math> to get that <math>2^8 + 2^5 \cdot 2^b + 2^{2b} = m^2</math>. To make the middle term have an exponent of <math>11</math>, we must have <math>b = 6</math>. Then <math>n = 12</math> and <math>m = (2^4 + 2^6)^2 = (16 + 64)^2 = 80^2</math>, so <math>m + n = \boxed{092}</math>. |
~ cxsmi | ~ cxsmi | ||
===Solution 3=== | ===Solution 3=== | ||
− | Calculating the terms on the RHS, we find that <math>256 + 2048 + 2^n = 2304 + 2^n = m^2</math>. We use trial-and-error to find a power of two that makes the RHS a perfect square. We find that <math>4096 = 2^12</math> works, and it produces <math>6400 = 80^2</math>. Then <math>m + n = \boxed{ | + | Calculating the terms on the RHS, we find that <math>256 + 2048 + 2^n = 2304 + 2^n = m^2</math>. We use trial-and-error to find a power of two that makes the RHS a perfect square. We find that <math>4096 = 2^12</math> works, and it produces <math>6400 = 80^2</math>. Then <math>m + n = \boxed{092}</math>. |
~ (also) cxsmi | ~ (also) cxsmi |
Revision as of 19:09, 4 February 2024
Contents
- 1 Introduction
- 2 User Counts
- 3 Cool asyptote graphs
- 4 Problems Sharing Contest
- 5 Contributions
- 6 Problems I made
- 7 Answer key
- 8 Solutions
- 8.1 Problem 1
- 8.2 Solution 1
- 8.3 Problem 2
- 8.4 Solution 1
- 8.5 Solution 2
- 8.6 Solution 3
- 8.7 Problem 3
- 8.8 Solution 1(Probably official MAA, lots of proofs)
- 8.9 Solution 2 (Fast, risky, no proofs)
- 8.10 Problem 4
- 8.11 Solution 1
- 8.12 Problem 5
- 8.13 Solution 1 (Euler's Totient Theorem)
- 8.14 Problem 6
- 8.15 Solution 1 (Recursion)
- 8.16 Problem 7
- 8.17 Solution 1
- 8.18 Problem 8
- 8.19 Solution 1
- 8.20 Problem 9
- 8.21 Solution 1(Wordless endless bash)
- 8.22 Problem 10
- 8.23 Solution 1 (Analytic geo)
- 8.24 Solution 2 (Hard vector bash)
- 9 Vandalism area
- 10 Problem 1
- 11 Problem 2
- 12 Problem 3
- 13 Problem 4
- 14 Problem 5
- 15 Problem 6
- 16 Problem 7
- 17 Problem 8
- 18 Problem 9
- 19 Problem 10
- 20 Problem 11
- 21 Problem 12
- 22 Problem 13
- 23 Problem 14
- 24 Problem 15
- 25 Problem 16
- 26 Problem 17
- 27 Problem 18
- 28 Problem 19
- 29 Problem 20
- 30 Problem 21
- 31 Problem 22
- 32 Problem 23
- 33 Problem 24
- 34 Problem 25
- 35 See Also
- 36 See also
Introduction
I am a 5th grader who likes making and doing problems, doing math, and redirecting pages (see Principle of Insufficient Reasons). I like geometry and don't like counting and probability. My number theory skill are also not bad.
User Counts
If this is your first time visiting this page, please change the number below by one. (Add 1, do NOT subtract 1)
For those of you who want more boxes, me too. However, this is the max number of boxes. Also, I check the pages history so I know if someone edited something.
(Please don't mess with the user count)
Doesn't that look like a number on a pyramid
Cool asyptote graphs
Asymptote is fun!
Problems Sharing Contest
Here, you can post all the math problem that you have. Everyone will try to come up with a appropriate solution. The person with the first solution will post the next problem. I'll start:
1. There is one and only one perfect square in the form
where and
are prime. Find that perfect square.
(DO NOT LOOK AT MY SOLUTIONS YET)
Contributions
2005 AMC 8 Problems/Problem 21 Solution 2
2022 AMC 12B Problems/Problem 25 Solution 5 (Now it's solution 6)
2023 AMC 12B Problems/Problem 20 Solution 3
2016 AIME I Problems/Problem 10 Solution 3
2017 AIME I Problems/Problem 14 Solution 2
2019 AIME I Problems/Problem 15 Solution 6
2022 AIME II Problems/Problem 3 Solution 3
Restored diagram for 1994 AIME Problems/Problem 7
Principle of Insufficient Reasons
Problems I made
Aime styled
Introductory
1. There is one and only one perfect square in the form
where and
are prime. Find that perfect square.
2. and
are positive integers. If
, find
.
Intermediate
3.The fraction,
where and
are side lengths of a triangle, lies in the interval
, where
and
are rational numbers. Then,
can be expressed as
, where
and
are relatively prime positive integers. Find
.
4. Suppose there is complex values and
that satisfy
Find .
5. Suppose
Find the remainder when is divided by
.
6. Suppose that there is rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are
other pegs positioned sufficiently apart. A
is made if
(i) ring changed position (i.e., that ring is transferred from one peg to another)
(ii) No rings are on top of smaller rings.
Then, let be the minimum possible number
that can transfer all
rings onto the second peg. Find the remainder when
is divided by
.
7. Suppose is a
-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are
roots, say
. Suppose all integers
ranging from
to
satisfies
. Also, suppose that
for an integer . If
is the minimum possible positive integral value of
.
Find the number of factors of the prime in
.
Olympiad
8. (Much harder) is an isosceles triangle where
. Let the circumcircle of
be
. Then, there is a point
and a point
on circle
such that
and
trisects
and
, and point
lies on minor arc
. Point
is chosen on segment
such that
is one of the altitudes of
. Ray
intersects
at point
(not
) and is extended past
to point
, and
. Point
is also on
and
. Let the perpendicular bisector of
and
intersect at
. Let
be a point such that
is both equal to
(in length) and is perpendicular to
and
is on the same side of
as
. Let
be the reflection of point
over line
. There exist a circle
centered at
and tangent to
at point
.
intersect
at
. Now suppose
intersects
at one distinct point, and
, and
are collinear. If
, then
can be expressed in the form
, where
and
are not divisible by the squares of any prime. Find
.
Someone mind making a diagram for this?
9. Suppose where
and
are relatively prime positive integers. Find
.
Proofs
10. In with
,
is the foot of the perpendicular from
to
.
is the foot of the perpendicular from
to
.
is the midpoint of
. Prove that
is perpendicular to
.
I will leave a big gap below this sentence so you won't see the answers accidentally.
Answer key
1. 049
2. 019
3. 092
4. 170
5. 736
6. 895
7. 011
8. 054
9. 077
Solutions
- Note: All the solutions so far have been made by me :)
Problem 1
There is one and only one perfect square in the form
where and
is prime. Find that perfect square.
Solution 1
.
Suppose
.
Then,
, so since ,
so
is less than both
and
and thus we have
and
. Adding them gives
so by Simon's Favorite Factoring Trick,
in some order. Hence,
.
Problem 2
and
are positive integers. If
, find
.
Solution 1
Let and
. Then,
Solution 2
Recall that a perfect square can be written as
. Since
is a perfect square, the
must be in this form. We substitute
for
to get that
. To make the middle term have an exponent of
, we must have
. Then
and
, so
.
~ cxsmi
Solution 3
Calculating the terms on the RHS, we find that . We use trial-and-error to find a power of two that makes the RHS a perfect square. We find that
works, and it produces
. Then
.
~ (also) cxsmi
Problem 3
The fraction,
where and
are side lengths of a triangle, lies in the interval
, where
and
are rational numbers. Then,
can be expressed as
, where
and
are relatively prime positive integers. Find
.
Solution 1(Probably official MAA, lots of proofs)
Lemma 1:
Proof: Since the sides of triangles have positive length, . Hence,
, so now we just need to find .
Since by the Trivial Inequality, we have
as desired.
To show that the minimum value is achievable, we see that if ,
, so the minimum is thus achievable.
Thus, .
Lemma 2:
Proof: By the Triangle Inequality, we have
.
Since , we have
.
Add them together gives
Even though unallowed, if , then
, so
.
Hence, , since by taking
and
close
, we can get
to be as close to
as we wish.
Solution 2 (Fast, risky, no proofs)
By the Principle of Insufficient Reason, taking we get either the max or the min. Testing other values yields that we got the max, so
. Another extrema must occur when one of
(WLOG,
) is
. Again, using the logic of solution 1 we see
so
so our answer is
.
Problem 4
Suppose there are complex values and
that satisfy
Find .
Solution 1
To make things easier, instead of saying , we say
.
Now, we have
.
Expanding gives
.
To make things even simpler, let
, so that .
Then, if , Newton's Sums gives
Therefore,
Now, we plug in
.
We substitute to get
.
Note: If you don't know Newton's Sums, you can also use Vieta's Formulas to bash.
Problem 5
Suppose
Find the remainder when is divided by 1000.
Solution 1 (Euler's Totient Theorem)
We first simplify
so
.
where the last step of all 3 congruences hold by the Euler's Totient Theorem. Hence,
Now, you can bash through solving linear congruences, but there is a smarter way. Notice that , and
. Hence,
, so
. With this in mind, we proceed with finding
.
Notice that and that
. Therefore, we obtain the system of congruences :
.
Solving yields , and we're done.
Problem 6
Suppose that there is rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are
other pegs positioned sufficiently apart. A
is made if
(i) ring changed position (i.e., that ring is transferred from one peg to another)
(ii) No bigger rings are on top of smaller rings.
Then, let be the minimum possible number
that can transfer all
rings onto the second peg. Find the remainder when
is divided by
.
Solution 1 (Recursion)
Let be the minimum possible number
that can transfer
rings onto the second peg. To build the recursion, we consider what is the minimum possible number
that can transfer
rings onto the second peg. If we use only legal
, then
will be smaller on the top, bigger on the bottom. Hence, the largest ring have to be at the bottom of the second peg, or the largest peg will have nowhere to go. In order for the largest ring to be at the bottom, we must first move the top
rings to the third peg using
, then place the largest ring onto the bottom of the second peg using
, and then get all the rings from the third peg on top of the largest ring using another
. This gives a total of
, hence we have
. Obviously,
. We claim that
. This is definitely the case for
. If this is true for
, then
so this is true for . Therefore, by induction,
is true for all
. Now,
. Therefore, we see that
.
But the part is trickier. Notice that by the Euler's Totient Theorem,
so is equivalent to the inverse of
in
, which is equivalent to the inverse of
in
, which, by inspection, is simply
. Hence,
, so by the Chinese Remainder Theorem, .
Problem 7
Suppose is a
-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are
roots, say
. Suppose all integers
ranging from
to
satisfies
. Also, suppose that
for an integer . If
is the minimum possible positive integral value of
.
Find the number of factors of the prime in
.
Solution 1
Since all integers ranging from
to
satisfies
, we have that all integers
ranging from
to
satisfies
, so by the Factor Theorem,
.
since is a
-degrees polynomial, and we let
to be the leading coefficient of
.
Also note that since is the roots of
,
Now, notice that
Similarly, we have
To minimize this, we minimize . The minimum
can get is when
, in which case
, so there is factors of
.
Problem 8
is an isosceles triangle where
. Let the circumcircle of
be
. Then, there is a point
and a point
on circle
such that
and
trisects
and
, and point
lies on minor arc
. Point
is chosen on segment
such that
is one of the altitudes of
. Ray
intersects
at point
(not
) and is extended past
to point
, and
. Point
is also on
and
. Let the perpendicular bisector of
and
intersect at
. Let
be a point such that
is both equal to
(in length) and is perpendicular to
and
is on the same side of
as
. Let
be the reflection of point
over line
. There exist a circle
centered at
and tangent to
at point
.
intersect
at
. Now suppose
intersects
at one distinct point, and
, and
are collinear. If
, then
can be expressed in the form
, where
and
are not divisible by the squares of any prime. Find
.
Someone mind making a diagram for this?
Solution 1
Line is tangent to
with point of tangency point
because
and
is perpendicular to
so this is true by the definition of tangent lines. Both
and
are on
and line
, so
intersects
at both
and
, and since we’re given
intersects
at one distinct point,
and
are not distinct, hence they are the same point.
Now, if the center of tangent circles are connected, the line segment will pass through the point of tangency. In this case, if we connect the center of
tangent circles,
and
(
and
respectively), it is going to pass through the point of tangency, namely,
, which is the same point as
, so
,
, and
are collinear. Hence,
and
are on both lines
and
, so
passes through point
, making
a diameter of
.
Now we state a few claims :
Claim 1: is equilateral.
Proof:
where the last equality holds by the Power of a Point Theorem.
Taking the square root of each side yields .
Since, by the definition of point ,
is on
. Hence,
, so
, and since
is the reflection of point
over line
,
, and since
, by the Pythagorean Theorem we have
Since is the perpendicular bisector of
,
and we have
hence
is equilateral.
With this in mind, we see that
Here, we state another claim :
Claim 2 : is a diameter of
Proof: Since , we have
and the same reasoning with gives
since
.
Now, apply Ptolemy’s Theorem gives
so is a diameter.
From that, we see that , so
. Now,
, so
, so
, and we’re done.
Problem 9
Suppose where
and
are relatively prime positive integers. Find
.
Solution 1(Wordless endless bash)
Problem 10
In with
,
is the foot of the perpendicular from
to
.
is the foot of the perpendicular from
to
.
is the midpoint of
. Prove that
.
Solution 1 (Analytic geo)
Let
We set it this way to simplify calculation when we calculate the coordinates of and
(Notice to find
, you just need to take the x coordinate of
and let the y coordinate be
).
Obviously,
Now, we see that
, so , as desired.
Solution 2 (Hard vector bash)
Solution 2a (Hard)
Hence, .
Solution 2b (Harder)
Since is the midpoint of
,
Now come the coordinates. Let
so that
.
Therefore,
Hence, we have that is perpendicular to
.
Vandalism area
Here, you can add anything, delete anything, and do anything! (Don't delete this line since it's instruction and don't be inappropriate) However, do NOT vandalize before this word (Feel free to delete this and the period that follows).
EX:
2021 Fall AMC 10A (Answer Key) Printable versions: • Fall AoPS Resources • Fall PDF | ||
Instructions
| ||
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 |
Problem 1
What is the value of ?
Problem 2
Menkara has a index card. If she shortens the length of one side of this card by
inch, the card would have area
square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by
inch?
Problem 3
What is the maximum number of balls of clay of radius that can completely fit inside a cube of side length
assuming the balls can be reshaped but not compressed before they are packed in the cube?
Problem 4
Mr. Lopez has a choice of two routes to get to work. Route A is miles long, and his average speed along this route is
miles per hour. Route B is
miles long, and his average speed along this route is
miles per hour, except for a
-mile stretch in a school zone where his average speed is
miles per hour. By how many minutes is Route B quicker than Route A?
Problem 5
The six-digit number is prime for only one digit
What is
Problem 6
Elmer the emu takes equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in
equal leaps. The telephone poles are evenly spaced, and the
st pole along this road is exactly one mile (
feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
Problem 7
As shown in the figure below, point lies on the opposite half-plane determined by line
from point
so that
. Point
lies on
so that
, and
is a square. What is the degree measure of
?
Problem 8
A two-digit positive integer is said to be if it is equal to the sum of its nonzero tens digit and the square of its units digit. How many two-digit positive integers are cuddly?
Problem 9
When a certain unfair die is rolled, an even number is times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even?
Problem 10
A school has students and
teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are
and
. Let
be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let
be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is
?
Problem 11
Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts
steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?
Problem 12
The base-nine representation of the number is
What is the remainder when
is divided by
Problem 13
Each of balls is randomly and independently painted either black or white with equal probability. What is the probability that every ball is different in color from more than half of the other
balls?
Problem 14
How many ordered pairs of real numbers satisfy the following system of equations?
Problem 15
Isosceles triangle has
, and a circle with radius
is tangent to line
at
and to line
at
. What is the area of the circle that passes through vertices
,
, and
Problem 16
The graph of is symmetric about which of the following? (Here
is the greatest integer not exceeding
.)
Problem 17
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at
,
, and
are
,
, and
meters, respectively. What is the height, in meters, of the pillar at
?
Problem 18
A farmer's rectangular field is partitioned into by
grid of
rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field?
Problem 19
A disk of radius rolls all the way around the inside of a square of side length
and sweeps out a region of area
. A second disk of radius
rolls all the way around the outside of the same square and sweeps out a region of area
. The value of
can be written as
, where
, and
are positive integers and
and
are relatively prime. What is
?
Problem 20
For how many ordered pairs of positive integers does neither
nor
have two distinct real solutions?
Problem 21
Each of the balls is tossed independently and at random into one of the
bins. Let
be the probability that some bin ends up with
balls, another with
balls, and the other three with
balls each. Let
be the probability that every bin ends up with
balls. What is
?
Problem 22
Inside a right circular cone with base radius and height
are three congruent spheres with radius
. Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is
?
Problem 23
For each positive integer , let
be twice the number of positive integer divisors of
, and for
, let
. For how many values of
is
Problem 24
Each of the edges of a cube is labeled
or
. Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the
faces of the cube equal to
?
Problem 25
A quadratic polynomial with real coefficients and leading coefficient is called
if the equation
is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial
for which the sum of the roots is maximized. What is
?
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by 2021 AMC 10B |
Followed by 2021 Fall AMC 10B | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.