Problems Collection
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
- 1 Contributed Problems and Solutions
- 2 AMC styled
- 3 AIME styled
- 4 Proofs
- 5 Solutions
- 5.1 Problem 1
- 5.2 Solution 1
- 5.3 Problem 2
- 5.4 Solution 1 (Slow, probably official MAA)
- 5.5 Solution 2 (Fast)
- 5.6 Solution 3 (Faster)
- 5.7 Problem 3
- 5.8 Solution 1(Probably official MAA, lots of proofs)
- 5.9 Solution 2 (Fast, risky, no proofs)
- 5.10 Problem 4
- 5.11 Solution 1
- 5.12 Problem 5
- 5.13 Solution 1 (Euler's Totient Theorem)
- 5.14 Problem 6
- 5.15 Solution 1 (Recursion)
- 5.16 Problem 7
- 5.17 Solution 1 (Tedious Casework)
- 5.18 Solution 2 (Official)
- 5.19 Solution 3 (Official and Fastest)
- 5.20 Problem 8
- 5.21 Solution 1
- 5.22 Problem 9
- 5.23 Solution 1
- 5.24 Problem 10
- 5.25 Solution 1(Wordless endless bash)
- 5.26 Problem 11
- 5.27 Solution 1 (Analytic geo)
- 5.28 Solution 2 (Hard vector bash)
- 6 Problems I made
- 7 Solutions
- 7.1 Problem 1
- 7.2 Solution 1
- 7.3 Problem 2
- 7.4 Solution 1 (Slow, probably official MAA)
- 7.5 Solution 2 (Fast)
- 7.6 Solution 3 (Faster)
- 7.7 Problem 3
- 7.8 Solution 1(Probably official MAA, lots of proofs)
- 7.9 Solution 2 (Fast, risky, no proofs)
- 7.10 Solution 3
- 7.11 Problem 4
- 7.12 Solution 1
- 7.13 Problem 5
- 7.14 Solution 1 (Euler's Totient Theorem)
- 7.15 Problem 6
- 7.16 Solution 1 (Recursion)
- 7.17 Problem 7
- 7.18 Solution 1
- 7.19 Problem 8
- 7.20 Solution 1
- 7.21 Problem 9
- 7.22 Solution 1(Wordless endless bash)
- 7.23 Problem 10
- 7.24 Solution 1 (Analytic geo)
- 7.25 Solution 2 (Hard vector bash)
- 8 Other Problems (This section will be deleted once I cleared out the page)
- 9 Contributors
- 10 See also
Contributed Problems and Solutions
WARNING: This page is going to be a very very messy page and will be hard to clear out!
AMC styled
AIME styled
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
.
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
ring changed position (i.e., that ring is transferred from one peg to another)
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. Let be a 2-digit positive integer satisfying
. Find the sum of all possible values of
.
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
.
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?
10. Suppose where
and
are relatively prime positive integers. Find
.
Proofs
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
is perpendicular to
.
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
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.
Solutions
I like your solutions.~Ddk001
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 (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
.
Solution 3
Expand the denominator. We now have . Consider its reciprocal; if this expression takes values on the interval
, then its reciprocal will take values on the interval
. This is important because we can now write the reciprocal of the expression as
. We attempt to maximize and minimize
. To maximize the expression, we consider the triangle inequality. From it, we find the following.
We rewrite.
Add all of the inequalities. We find the following.
Considering the equality case and plugging into the expression, we find that the maximum value of the expression is
. However, since this "equality case" cannot actually happen, this part of the interval must be open. Now, we minimize the inequality by using the Power Mean Inequality (specifically, the QM-AM part of the inequality). Considering the terms
,
, and
, we find the following.
Square both sides.
Rewrite as follows.
Considering the equality case and plugging into the expression, we find that the minimum value of the expression is
. Since the expression (which we said was the reciprocal of the original expression) takes values on the interval
, the original expression must take values on the interval
. Then
, so our final answer is
.
~ cxsmi
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
.
Other Problems (This section will be deleted once I cleared out the page)
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 .
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.