Difference between revisions of "Problems Collection"
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Please contribute whatever problems you have. | Please contribute whatever problems you have. | ||
+ | ==Problems== | ||
+ | ===AMC styled=== | ||
+ | Nothing yet to show | ||
+ | ===AIME styled=== | ||
+ | 1.There is one and only one perfect square in the form | ||
+ | <cmath>(p^2+1)(q^2+1)-((pq)^2-pq+1)</cmath> | ||
+ | |||
+ | where <math>p</math> and <math>q</math> is prime. Find that perfect square. | ||
+ | |||
+ | |||
+ | 2.<math>m</math> and <math>n</math> are positive integers. If <math>m^2=2^8+2^{11}+2^n</math>, find <math>m+n</math>. | ||
+ | |||
+ | |||
+ | 3.The fraction, | ||
+ | |||
+ | <cmath>\frac{ab+bc+ac}{(a+b+c)^2}</cmath> | ||
+ | |||
+ | where <math>a,b</math> and <math>c</math> are side lengths of a triangle, lies in the interval <math>(p,q]</math>, where <math>p</math> and <math>q</math> are rational numbers. Then, <math>p+q</math> can be expressed as <math>\frac{r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r+s</math>. | ||
+ | |||
+ | |||
+ | 4. | ||
+ | ===Olympaid styled=== | ||
+ | ===Putnam styled (Calculus version of AMC, AIME, and Olympaid)=== | ||
+ | ===Others (proofs & ect.)=== | ||
==Solutions== | ==Solutions== | ||
===Problem 1=== | ===Problem 1=== |
Revision as of 19:50, 2 August 2024
This is a page where you can share the problems you made (try not to use past exams).
If you have problems or solutions to problems already on this page to contribute, please put it below. In the former case, please give at least one solution to that problem. Problems/solutions in the following section shall be inspected by Ddk001 and will be put in the section after that section. If you would like, put your user name to the list of contributors of this page (at the near-end).
Contents
[hide]- 1 Contributed Problems and Solutions
- 2 Problems
- 3 Solutions
- 3.1 Problem 1
- 3.2 Solution 1
- 3.3 Problem 2
- 3.4 Solution 1 (Slow, probably official MAA)
- 3.5 Solution 2 (Fast)
- 3.6 Solution 3 (Faster)
- 3.7 Problem 3
- 3.8 Solution 1(Probably official MAA, lots of proofs)
- 3.9 Solution 2 (Fast, risky, no proofs)
- 3.10 Problem 4
- 3.11 Solution 1
- 3.12 Problem 5
- 3.13 Solution 1 (Euler's Totient Theorem)
- 3.14 Problem 6
- 3.15 Solution 1 (Recursion)
- 3.16 Problem 7
- 3.17 Solution 1 (Tedious Casework)
- 3.18 Solution 2 (Official)
- 3.19 Solution 3 (Official and Fastest)
- 3.20 Problem 8
- 3.21 Solution 1
- 3.22 Problem 9
- 3.23 Solution 1
- 3.24 Problem 10
- 3.25 Solution 1(Wordless endless bash)
- 3.26 Problem 11
- 3.27 Solution 1 (Analytic geo)
- 3.28 Solution 2 (Hard vector bash)
- 4 Other Problems
- 5 Contributors
- 6 See also
Contributed Problems and Solutions
Please contribute whatever problems you have.
Problems
AMC styled
Nothing yet to show
AIME styled
1.There is one and only one perfect square in the form
where and is prime. Find that perfect square.
2. and are positive integers. If , find .
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.
Olympaid styled
Putnam styled (Calculus version of AMC, AIME, and Olympaid)
Others (proofs & ect.)
Solutions
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, .~Ddk001
Problem 2
and are positive integers. If , find .
Solution 1 (Slow, probably official MAA)
Let and . Then,
Solution 2 (Fast)
Recall that a perfect square can be written as . Since is a perfect square, the RHS 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 (Faster)
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 . ~Ddk001
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.~Ddk001
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. ~Ddk001
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
Let be a 2-digit positive integer satisfying . Find the sum of all possible values of .
Solution 1 (Tedious Casework)
Case 1:
In this case, we have
.
If , we must have
, but this contradicts the original assumption of , so hence we must have .
With this in mind, we consider the unit digit of .
Subcase 1.1:
In this case, we have that
.
There is no apparent contradiction here, so we leave this as it is.
Subcase 1.2:
In this case, we have that
.
This contradicts with the fact that , so this is impossible.
Subcase 1.3:
In this case, we have that
.
However, this is impossible for all .
Subcase 1.4:
In this case, we have that
.
Again, this yields , which, again, contradicts .
Hence, we must have .
Now, with determined by modular arithmetic, we actually plug in the values.
To simplify future calculations, note that
.
For , this does not hold.
For , this does not hold.
For , this does hold. Hence, is a solution.
For , this does not hold.
For , this does not hold.
Hence, there is no positive integers and between and inclusive such that .
Case 2:
For this case, we must have
which is impossible if a is a integer and .
Case 3:
In this case, we have
.
If , we must have
which is impossible since and .
Hence, .
Subcase 3.1:
Testing cases, we can see that there is no such .
Subcase 3.2:
Testing cases, we can see that there is no such .
Subcase 3.3:
Testing cases, we can see that there is no such .
Subcase 3.4:
Testing cases, we can see that there is no such .
We see that . ~Ddk001
Solution 2 (Official)
cleary square of a positive double-digit integer is either 3-digit or 4-digit as it ranges between and .
This means both and are less than or equal to 7 as any greater than 7! exceeds 9999.
Now at least one of or must be greater than or equal to 5 or else the maximum possible value of LHS is 4!+4!=48<100 also, both and can't be greater than or equal to 5 as if they were so. The unit digit of LHS would be zero but the unit digit of RHS would be either 5,6 or 9.
Hence, one of and is greater than or equal to 5 and the other is less than 5. this means the unit digit of RHS is the unit digit of a factorial which is less than 5.
Hence unit digit of RHS is 0,1,2, 6 or 4. 0,2,4 and 6 are rejected as follows:-
1. 2 can't be the unit digit of a perfect square.
2. If 6 is the unit digit of OF LHS this means one of the numbers is 3( as only 3! has the unit digit 6) and also this would mean that is either 4 or 6. This directly means that 36 and 34 are the only cases to be tested. 34 can't be as both the digits are less than 5 and 36 clearly doesn't satisfy
3. If 0 is the unit digit of LHS then 50 60 70 are the only cases (as one of the digits is greater than or equal to 5) that don't satisfy
4. If 4 is the unit digit of LHS this means one of the numbers is 4( as only 4! has the unit digit 4) and also this would mean that is either 2 or 8. This directly means that 42 and 48 are the only cases to be tested. 42 can't be as both the digits are less than 5 and 48 can't also as one of the digits is 8
Hence, the unit digit of LHS must be 1 for which =1 or 9(rejected). This means 51 61 and 71 are the only cases to be tried now which is really very easy to calculate and get 71 as the only possible solution to the problem and
thus, our answer is 7+1=.~ SANSGANKRSNGUPTA
Solution 3 (Official and Fastest)
cleary square of a positive double-digit integer is either 3-digit or 4-digit as it ranges between and .
This means both and are less than or equal to 7 as any greater than 7! exceeds 9999.
Now at least one of or must be greater than or equal to 5 or else the maximum possible value of LHS is 4!+4!=48<100 also, both and can't be greater than or equal to 5 as if they were so. The unit digit of LHS would be zero but the unit digit of RHS would be either 5,6 or 9.
Hence, one of and is greater than or equal to 5 and the other is less than 5. Also, RHS is a perfect square, so it must be 0 or 1
Hence the number less than 5 can be either 1 or 4 because 2! and 3! are 2
If it is 4, then the unit digit of RHS is 4 meaning is either 2 or 8 both of which aren't possible as 8>7 and the number less than 5 is 4 not 2.
If it is 1 then the unit digit of RHS is 1 implying is either 9(rejected 9>7) or 1. This means is 1, this means 51 61 and 71 are the only cases to be tried
which is very easy to calculate and get 71 as the only possible solution to the problem and
thus, our answer is 7+1=.~ SANSGANKRSNGUPTA
Problem 8
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 . ~Ddk001
Problem 9
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. ~Ddk001
Problem 10
Suppose where and are relatively prime positive integers. Find .
Solution 1(Wordless endless bash)
Problem 11
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. ~Ddk001
Solution 2 (Hard vector bash)
Solution 2a (Hard)
Hence, . ~Ddk001
Solution 2b (Harder)
Since is the midpoint of ,
Now come the coordinates. Let
so that
.
Therefore,
Hence, we have that is perpendicular to . ~Ddk001
Other Problems
Beware! Below shall be a very long solution (and, of course, its corresponding problem).
Problem: Show that the series
where p and m are real numbers converge if or but and diverge otherwise.
Solution:
Before we even get started, let's state a few definitions first.
Definition 1: A set , whose elements we shall call points, is said to be a metric space if with any two points and of there is associated a real number , called the distance from to , such that
if
, for any .
Any function with these properties is called a distance function or a metric
Definition 2: Let be a metric space. All points and sets mentioned below are understood to be elements and subsets of .
(a) A neighborhood of a point is a set consisting of all points such that . The number is said to be the radius of .
(b) A point is a limit point of a set if every neighborhood of contains a point such that .
(c) If and is not a limit point of , then is called a isolated point of .
(d) is closed if every limit point of is a point of .
(e) A point is an interior point of if there is a neighborhood of such that
(f) is open if every point of is a interior point of .
(g) The complement of (denoted by ) is the set of points such that but .
(h) is bounded if there is a ream number and a point such that for all .
Definition 3: Let be a metric space. All points and sets mentioned below are understood to be elements and subsets of .
(a) By a open cover of a set in we mean a collection of open subsets of such that .
(b) A subset of is said to be compact if every open cover of contain a finite subcover.
Definition 4: Let be a metric space and . Let denote the set of limit points of . The closure of is said to be .
Lemma 1: Closed subsets of compact sets are compact.
Proof: Suppose , is closed and is compact. Let be a open cover of . We wish to show that have no finite subcover of . Let be adjoined to . Then we obtain an open cover of . (Note that complements of closed sets are open since if is closed, all the limit points of is a point of so if , . Every neighborhood of contain a point of that is not itself so no neighborhood of is completely in so is not a interior point of , so is open. Thus is open.) Since is compact, there is a finite subcollection of that cover , and hence . If , we can exclude it from and can thus still have a open cover of . The obtained open cover is thus a finite subcollection of that cover , so is compact.
Definition 5: A subset of is said to be a k-cell if is the set of such that , where the 's and 's are real numbers.
Lemma 2: Every k-cell is compact.
Proof: Let be a k-cell, consisting of the points such that . Put
Thus would imply .
Suppose, for the sake of contradiction, that there exist an open cover of which contain no finite subcover. Put . The intervals and then determine k-cells whose union is . Thus at least one of these sets, call it , cannot be covered by any finite subcollection of (otherwise could be so covered). We divide like we did , and obtain as we did . As the process continue we thus obtain a sequence of 's such that
(a) ;
(b) is not covered by any subcollection of ;
(c) If , then .
We claim that there is a point in every . In fact, we can even prove the stronger statement
"Let be a positive interger. If is a sequence of k-cells such that , then ."
To prove that, it suffices to prove it with . The rest follows easily from induction. The case can be rephrased as follows:
"If is a sequence of intervals such that , then
To prove this, let . Then let denote the set of all the 's. Obviously is non-empty and is bounded above (by , for one). Then the least upper bound of exist, call it . If and are any positive integers, then
so for each . Since , obviously .
The result follow.
Thus there is a point in every . Now, since covers it follows that for some . Since is open there is a such that implies . If n is so large that then (c) implies that , which contradicts (b). A contradiction is obtained and the result must follow.
Corollary 1 (Heine-Borel Theorem): Suppose . Then the following two statements are equivalent:
(a) is closed and bounded
(b) is compact
Proof: It suffices to show that . We shall only prove that since we will only use that part. However, a proof of the converse is given in a remark after the solution.
Since is closed and bounded, is a closed subset of a k-cell, a compact set. Thus is compact by Lemma 1.
Thus, in , every single bounded set have a compact closure.
Definition 6: Let be a subset of a metric space , and let be the set such that
.
Then the smallest upper bound of is called the diameter of and is often written as .
Lemma 3: (a) Let be the closure of in a metric space . Then
(b) If is a sequence of compact sets in such that and if
, then consist of exactly one point.
Proof: (a) Since , it is obvious that
Thus it suffice to show that
To do so, choose . Then for each there exist points and such that and that . Hence,
It follows that
. Since is ambitarily chosen, (a) is proved.
(b) Put . It suffice to show that has at most 1 point and that it exist. The former is easy. If have 1 or more points, which contradict the fact that .
It remains to show that is non-empty.
Sublemma 3.1: If is a collection of compact subsets of a metric space such that the intersection of every finite subcollection of is non-empty, then is non-empty.
Proof: Fix a member of and let for each . Suppose, for the sake of contradiction that no point of belong to every single member of . Note that every is open by the parenthetical remark in the proof of lemma 1. Thus the sets form a open cover of ; however, since is compact there exist finitely many indices such that
But this would mean
, contradictory to the hypothesis.
A contradiction is obtained so we are done. .
The result follow from the sublemma.
Now, we show state another important lemma that will use the above lemma (lemma 3). Part (a) and (c) of the following lemma is called the Cauchy Criterion.
Before starting the proof, however, we have one more definition to make.
Definition 7: A sequence of a metric space is said to be a Cauchy Sequence if for every there exist a integer such that if .
Obviously, we can state definition 7 in terms of definition 6:
" A sequence of a metric space is said to be a Cauchy Sequence if
where is the set consisting of the points . "
In fact, its converse is also true (and its proof is trivial):
"Let be a sequence of sets such that
.
Then if a sequence is chosen with
Contributors
See also
Here's the source for the problems:
1,2,3,4,5,6,8,9,10,11: Ddk001, credits given to Ddk001
7: SANSKAR'S OG PROBLEMS, credits given to SANSGANKRSNGUPTA
- Note: Problem 6 is based on the Tower of Hanoi Problem
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.