Quadratic polynomials and
have leading coefficients
and
respectively. The graphs of both polynomials pass through the two points
and
Find
Let Since the
-terms of
and
cancel, we conclude that
is a linear polynomial.
Note thatso the slope of
is
It follows that the equation of is
for some constant
and we wish to find
We substitute into this equation to get
from which
~MRENTHUSIASM
Letfor some constants
and
We are given thatand we wish to find
We need to cancel
and
Since
we subtract
from
to get
~MRENTHUSIASM
Let
By substitutes and
into these equations, we can get:
Hence,
and
.
Similarly,Hence,
and
.
Notice that and
. Therefore
~Littlemouse
Find the three-digit positive integer whose representation in base nine is
where
and
are (not necessarily distinct) digits.
We are given thatwhich rearranges to
Taking both sides modulo
we have
The only solution occurs at
from which
Therefore, the requested three-digit positive integer is
~MRENTHUSIASM
As shown in Solution 1, we get .
Note that and
are large numbers comparatively to
, so we hypothesize that
and
are equal and
fills the gap between them. The difference between
and
is
, which is a multiple of
. So, if we multiply this by
, it will be a multiple of
and thus the gap can be filled. Therefore, the only solution is
, and the answer is
.
~KingRavi
As shown in Solution 1, we get
We list a few multiples of out:
Of course,
can't be made of just
's. If we use one
, we get a remainder of
, which can't be made of
's either. So
doesn't work.
can't be made up of just
's. If we use one
, we get a remainder of
, which can't be made of
's. If we use two
's, we get a remainder of
, which can be made of
's. Therefore we get
so
and
. Plugging this back into the original problem shows that this answer is indeed correct. Therefore,
~Technodoggo
As shown in Solution 1, we get .
We can see that is
larger than
, and we have an
. We can clearly see that
is a multiple of
, and any larger than
would result in
being larger than
. Therefore, our only solution is
. Our answer is
.
~Arcticturn
In isosceles trapezoid , parallel bases
and
have lengths
and
, respectively, and
. The angle bisectors of
and
meet at
, and the angle bisectors of
and
meet at
. Find
.
~MRENTHUSIASM ~ihatemath123
We have the following diagram:
Let and
be the points where
and
extend to meet
, and
be the height of
. As proven in Solution 2, triangles
and
are congruent right triangles. Therefore,
. We can apply this logic to triangles
and
as well, giving us
. Since
,
.
Additionally, we can see that is similar to
and
. We know that
. So, we can say that the height of the triangle
is
while the height of the triangle
is
. After that, we can figure out the distance from
to
and the height of triangle
.
Finally, since the ratio between the height of to the height of
is
and
is
,
~Cytronical
Extend line to meet
at
and
at
. The diagram looks like this:
Because the trapezoid is isosceles, by symmetry
is parallel to
and
. Therefore,
by interior angles and
by the problem statement. Thus,
is isosceles with
. By symmetry,
is also isosceles, and thus
. Similarly, the same thing is happening on the right side of the trapezoid, and thus
is the midline of the trapezoid. Then,
.
Since and
, we have
. The length of the midline of a trapezoid is the average of their bases, so
. Finally,
.
~KingRavi
We have the following diagram:Extend lines
and
to meet line
at points
and
, respectively, and extend lines
and
to meet
at points
and
, respectively.
Claim: quadrilaterals and
are rhombuses.
Proof: Since ,
. Therefore, triangles
,
,
and
are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence,
is congruent to the other three. Therefore,
, so
is a rhombus. By symmetry,
is also a rhombus.
Extend line to meet
and
at
and
, respectively. Because of rhombus properties,
. Also, by rhombus properties,
and
are the midpoints of segments
and
, respectively; therefore, by trapezoid properties,
. Finally,
.
~ihatemath123
Let and
be the feet of the altitudes from
and
, respectively, to
, and let
and
be the feet of the altitudes from
and
, respectively, to
. Side
is parallel to side
, so
is a rectangle with width
. Furthermore, because
and trapezoid
is isosceles,
.
Also because is isosceles,
is half the total sum of angles in
, or
. Since
and
bisect
and
, respectively, we have
, so
.
Letting , applying Pythagoras to
yields
. We then proceed using similar triangles:
and
, so by AA similarity
. Likewise,
and
, so by AA similarity
. Thus
.
Adding our two equations for and
gives
. Therefore, the answer is
.
~Orange_Quail_9
This will be my first solution on AoPS. My apologies in advance for any errors.
Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that is equidistant from
and
and
is equidistant from
and
If we let the feet of the altitudes from
to
and
be called
and
respectively, we can say that
Analogously, we let the feet of the altitudes from
to
and
be
and
respectively. Thus,
Because
is an isosceles trapezoid, we can say that all of the altitudes are equal to each other.
By SA as well as SS congruence for right triangles, we find that triangles and
are congruent. Similarly,
and
by the same reasoning. Additionally,
since
and
are congruent rectangles.
If we then let let
and let
we can create the following system of equations with the given side length information:
Adding the first two equations, subtracting by twice the second, and dividing by
yields
~regular
Let and
where
Find the number of ordered pairs
of positive integers not exceeding
that satisfy the equation
We rewrite and
in polar form:
The equation
becomes
for some integer
Since and
we conclude that
Note that the values for
and the values for
have one-to-one correspondence.
We apply casework to the values for
There are values for
so there are
values for
It follows that
so there are
values for
There are ordered pairs
in this case.
There are values for
so there are
values for
It follows that
so there are
values for
There are ordered pairs
in this case.
There are values for
so there are
values for
It follows that
so there are
values for
There are ordered pairs
in this case.
Together, the answer is
~MRENTHUSIASM
First we recognize that and
because the cosine and sine sums of those angles give the values of
and
, respectively. By Demoivre's theorem,
. When you multiply by
, we can think of that as rotating the complex number
counterclockwise in the complex plane. Therefore, by the equation we know that
and
land on the same angle.
This means thatwhich we can simplify to
Notice that this means that
cycles by
for every value of
. This is because once
hits
, we get an angle of
and the angle laps onto itself again. By a similar reasoning,
laps itself every
times, which is much easier to count. By listing the possible values out, we get the pairs
:
We have
columns in total:
values for the first column,
for the second,
for the third, and then
for the fourth,
for the fifth,
for the sixth, etc. Therefore, this cycle repeats every
columns and our total sum is
.
~KingRavi
A straight river that is meters wide flows from west to east at a rate of
meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of
meters downstream from Sherry. Relative to the water, Melanie swims at
meters per minute, and Sherry swims at
meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find
.
Define as the number of minutes they swim for.
Let their meeting point be . Melanie is swimming against the current, so she must aim upstream from point
, to compensate for this; in particular, since she is swimming for
minutes, the current will push her
meters downstream in that time, so she must aim for a point
that is
meters upstream from point
. Similarly, Sherry is swimming downstream for
minutes, so she must also aim at point
to compensate for the flow of the current.
If Melanie and Sherry were to both aim at point in a currentless river with the same dimensions, they would still both meet at that point simultaneously. Since there is no current in this scenario, the distances that Melanie and Sherry travel, respectively, are
and
meters. We can draw out this new scenario, with the dimensions that we have:
(While it is indeed true that the triangle above with side lengths
,
and
is a right triangle, we do not know this yet, so we cannot assume this based on the diagram.)
By Pythagorean, we have
Subtracting the first equation from the second gives us , so
. Substituting this into our first equation, we have that
So .
~ihatemath123
We have the following diagram:Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed.
Let and
be some positive numbers. We have the following table:
Recall that
so
We subtract
from
to get
from which
Substituting this into either equation, we have
It follows that Melanie and Sherry both swim for minutes. Therefore, the answer is
~MRENTHUSIASM
Find the number of ordered pairs of integers such that the sequence
is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
Since and
cannot be an arithmetic progression,
or
can never be
. Since
and
cannot be an arithmetic progression,
can never be
. Since
, there are
ways to choose
and
with these two restrictions in mind.
However, there are still specific invalid cases counted in these pairs
. Since
cannot form an arithmetic progression,
.
cannot be an arithmetic progression, so
; however, since this pair was not counted in our
, we do not need to subtract it off.
cannot form an arithmetic progression, so
.
cannot form an arithmetic progression, so
.
cannot form an arithmetic progression,
; however, since this pair was not counted in our
(since we disallowed
or
to be
), we do not to subtract it off.
Also, the sequences ,
,
,
,
and
will never be arithmetic, since that would require
and
to be non-integers.
So, we need to subtract off progressions from the
we counted, to get our final answer of
.
~ ihatemath123
We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.
We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50. (Note if we take only 1 at a time, there will have to be 3 of
, which is impossible.)
If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.
Finally, we count the sequences that are something like (one of 3,4,5,), , (one of 30, 40, 50). If this is to be the case, then let
be the starting value in the sequence. The sequence will be
; We see that if we subtract the largest term by the smallest term we have
, so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are
. Of these, only the last is invalid because it gives
, larger than our bounds
. Therefore, we subtract
from this case.
Our final answer is
~KingRavi
Denote .
Denote by a subset of
, such that there exists an arithmetic sequence that has 4 terms and includes
but not
.
Denote by a subset of
, such that there exists an arithmetic sequence that has 4 terms and includes
but not
.
Hence, is a subset of
, such that there exists an arithmetic sequence that has 4 terms and includes both
and
.
Hence, this problem asks us to compute
First, we compute .
We have .
Second, we compute .
:
.
We have . Thus, the number of solutions is 21.
:
.
We have . Thus, the number of solutions is 9.
Thus, .
Third, we compute .
In , we have
. However, because
, we have
. Thus,
.
This implies . Thus,
.
Fourth, we compute .
: In the arithmetic sequence, the two numbers beyond
and
are on the same side of
and
.
Hence, . Therefore, the number solutions in this case is 3.
: In the arithmetic sequence, the two numbers beyond
and
are on the opposite sides of
and
.
: The arithmetic sequence is
.
Hence, .
: The arithmetic sequence is
.
Hence, .
: The arithmetic sequence is
.
Hence, .
Putting two cases together, .
Therefore,
~Steven Chen (www.professorchenedu.com)
divide cases into .(Notice that
can't be equal to
, that's why I divide them into two parts. There are three cases that arithmetic sequence forms:
.(NOTICE that
IS NOT A VALID SEQUENCE!) So when
, there are
possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)
When , there are
ways.
In all, there are possible sequences.
~bluesoul
Find the number of ordered pairs of integers such that the sequence
is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
Since and
cannot be an arithmetic progression,
or
can never be
. Since
and
cannot be an arithmetic progression,
can never be
. Since
, there are
ways to choose
and
with these two restrictions in mind.
However, there are still specific invalid cases counted in these pairs
. Since
cannot form an arithmetic progression,
.
cannot be an arithmetic progression, so
; however, since this pair was not counted in our
, we do not need to subtract it off.
cannot form an arithmetic progression, so
.
cannot form an arithmetic progression, so
.
cannot form an arithmetic progression,
; however, since this pair was not counted in our
(since we disallowed
or
to be
), we do not to subtract it off.
Also, the sequences ,
,
,
,
and
will never be arithmetic, since that would require
and
to be non-integers.
So, we need to subtract off progressions from the
we counted, to get our final answer of
.
~ ihatemath123
We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing.
We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50. (Note if we take only 1 at a time, there will have to be 3 of
, which is impossible.)
If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}.
Finally, we count the sequences that are something like (one of 3,4,5,), , (one of 30, 40, 50). If this is to be the case, then let
be the starting value in the sequence. The sequence will be
; We see that if we subtract the largest term by the smallest term we have
, so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are
. Of these, only the last is invalid because it gives
, larger than our bounds
. Therefore, we subtract
from this case.
Our final answer is
~KingRavi
Denote .
Denote by a subset of
, such that there exists an arithmetic sequence that has 4 terms and includes
but not
.
Denote by a subset of
, such that there exists an arithmetic sequence that has 4 terms and includes
but not
.
Hence, is a subset of
, such that there exists an arithmetic sequence that has 4 terms and includes both
and
.
Hence, this problem asks us to compute
First, we compute .
We have .
Second, we compute .
:
.
We have . Thus, the number of solutions is 21.
:
.
We have . Thus, the number of solutions is 9.
Thus, .
Third, we compute .
In , we have
. However, because
, we have
. Thus,
.
This implies . Thus,
.
Fourth, we compute .
: In the arithmetic sequence, the two numbers beyond
and
are on the same side of
and
.
Hence, . Therefore, the number solutions in this case is 3.
: In the arithmetic sequence, the two numbers beyond
and
are on the opposite sides of
and
.
: The arithmetic sequence is
.
Hence, .
: The arithmetic sequence is
.
Hence, .
: The arithmetic sequence is
.
Hence, .
Putting two cases together, .
Therefore,
~Steven Chen (www.professorchenedu.com)
divide cases into .(Notice that
can't be equal to
, that's why I divide them into two parts. There are three cases that arithmetic sequence forms:
.(NOTICE that
IS NOT A VALID SEQUENCE!) So when
, there are
possible ways( 3 means the arithmetic sequence and 13 means there are 13 "a" s and b cannot be 20)
When , there are
ways.
In all, there are possible sequences.
~bluesoul
Let be distinct integers from
to
The minimum possible positive value of
can be written as
where
and
are relatively prime positive integers. Find
To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that
If we minimize the numerator, then Note that
so
It follows that
and
are consecutive composites with prime factors no other than
and
The smallest values for
and
are
and
respectively. So, we have
and
from which
If we do not minimize the numerator, then Note that
Together, we conclude that the minimum possible positive value of is
Therefore, the answer is
~MRENTHUSIASM ~jgplay
Equilateral triangle is inscribed in circle
with radius
Circle
is tangent to sides
and
and is internally tangent to
Circles
and
are defined analogously. Circles
and
meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of
are the vertices of a large equilateral triangle in the interior of
and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of
The side length of the smaller equilateral triangle can be written as
where
and
are positive integers. Find
~MRENTHUSIASM ~ihatemath123
We can extend and
to
and
respectively such that circle
is the incircle of
.
Since the diameter of the circle is the height of this triangle, the height of this triangle is
. We can use inradius or equilateral triangle properties to get the inradius of this triangle is
(The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is
.
Let be the center of the largest circle. We will set up a coordinate system with
as the origin. The center of
will be at
because it is directly beneath
and is the length of the larger radius minus the smaller radius, or
. By rotating this point
around
, we get the center of
. This means that the magnitude of vector
is
and is at a
degree angle from the horizontal. Therefore, the coordinates of this point are
and by symmetry the coordinates of the center of
is
.
The upper left and right circles intersect at two points, the lower of which is . The equations of these two circles are:
We solve this system by subtracting to get
. Plugging back in to the first equation, we have
. Since we know
is the lower solution, we take the negative value to get
.
We can solve the problem two ways from here. We can find by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find
as they lie on the same vertical,
is
degrees so we can make use of
triangles, and
because
is the center of triangle
. We can draw the diagram as such:
Note that
. It follows that
Finally, the answer is
.
~KingRavi
For equilateral triangle with side length
, height
, and circumradius
, there are relationships:
,
, and
.
There is a lot of symmetry in the figure. The radius of the big circle is
, let the radius of the small circles
,
,
be
.
We are going to solve this problem in steps:
We have is a
triangle, and
,
(
and
are tangent), and
. So, we get
and
.
Since and
are tangent, we get
.
Note that is an equilateral triangle, and
is its center, so
.
Note that is an isosceles triangle, so
In , Power of a Point gives
and
.
It follows that . We solve this quadratic equation:
.
Since is the circumradius of equilateral
, we have
.
Therefore, the answer is .
~isabelchen
Ellina has twelve blocks, two each of red (), blue (
), yellow (
), green (
), orange (
), and purple (
). Call an arrangement of blocks
if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement
is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is
where
and
are relatively prime positive integers. Find
Consider this position chart:Since there has to be an even number of spaces between each ball of the same color, spots
,
,
,
,
, and
contain some permutation of all 6 colored balls. Likewise, so do the even spots, so the number of even configurations is
(after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of
possible arrangements, so the probability is:
which is in simplest form. So,
.
~Oxymoronic15
We can simply use constructive counting. First, let us place the red balls; choose the first slot in ways, and the second in
ways, because the number is cut in half due to the condition in the problem. This gives
ways to place the blue balls. Similarly, there are
ways to place the blue balls, and so on, until there are
ways to place the purple balls. Thus, the probability is
and the desired answer extraction is
.
~A1001
Three spheres with radii ,
, and
are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at
,
, and
, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that
. Find
.
We let be the plane that passes through the spheres and
and
be the centers of the spheres with radii
and
. We take a cross-section that contains
and
, which contains these two spheres but not the third. Because the plane cuts out congruent circles, they have the same radius and from the given information,
. Since
is a trapezoid, we can drop an altitude from
to
to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is
and let the distance from
to
be
. Then we have
.
We have because of the rectangle, so
. Squaring, we have
. Subtracting, we get
. We also notice that since we had
means that
and since we know that
,
.
We now look at our second diagram.
. Since
, we have
. Using Pythagorean theorem,
. Therefore,
~KingRavi
Let the distance between the center of the sphere to the center of those circular intersections as separately.
. According to the problem, we have
. After solving we have
, plug this back to
The desired value is
~bluesoul
Denote by the radius of three congruent circles formed by the cutting plane. Denote by
,
,
the centers of three spheres that intersect the plane to get circles centered at
,
,
, respectively.
Because three spheres are mutually tangent, ,
.
We have ,
,
.
Because and
are perpendicular to the plane,
is a right trapezoid, with
.
Hence,
Recall that
Hence, taking , we get
Solving (1) and (3), we get and
.
Thus, .
Thus, .
Because and
are perpendicular to the plane,
is a right trapezoid, with
.
Therefore,
In our solution, we do not use the conditio that spheres
and
are externally tangent. This condition is redundant in solving this problem.
~Steven Chen (www.professorcheneeu.com)
Three spheres with radii ,
, and
are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at
,
, and
, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that
. Find
.
We let be the plane that passes through the spheres and
and
be the centers of the spheres with radii
and
. We take a cross-section that contains
and
, which contains these two spheres but not the third. Because the plane cuts out congruent circles, they have the same radius and from the given information,
. Since
is a trapezoid, we can drop an altitude from
to
to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is
and let the distance from
to
be
. Then we have
.
We have because of the rectangle, so
. Squaring, we have
. Subtracting, we get
. We also notice that since we had
means that
and since we know that
,
.
We now look at our second diagram.
. Since
, we have
. Using Pythagorean theorem,
. Therefore,
~KingRavi
Let the distance between the center of the sphere to the center of those circular intersections as separately.
. According to the problem, we have
. After solving we have
, plug this back to
The desired value is
~bluesoul
Denote by the radius of three congruent circles formed by the cutting plane. Denote by
,
,
the centers of three spheres that intersect the plane to get circles centered at
,
,
, respectively.
Because three spheres are mutually tangent, ,
.
We have ,
,
.
Because and
are perpendicular to the plane,
is a right trapezoid, with
.
Hence,
Recall that
Hence, taking , we get
Solving (1) and (3), we get and
.
Thus, .
Thus, .
Because and
are perpendicular to the plane,
is a right trapezoid, with
.
Therefore,
In our solution, we do not use the conditio that spheres
and
are externally tangent. This condition is redundant in solving this problem.
Let be a parallelogram with
. A circle tangent to sides
,
, and
intersects diagonal
at points
and
with
, as shown. Suppose that
,
, and
. Then the area of
can be expressed in the form
, where
and
are positive integers, and
is not divisible by the square of any prime. Find
.
Let's redraw the diagram, but extend some helpful lines.
We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let be our tangents from the circle to the parallelogram. By the secant power of a point, the power of
. Then
. Similarly, the power of
and
. We let
and label the diagram accordingly.
Notice that because . Let
be the center of the circle. Since
and
intersect
and
, respectively, at right angles, we have
is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from
to
and
to
, and both are equal to
. Since
,
. Since
and
. We can now use Pythagorean theorem on
; we have
and
.
We know that because
is a parallelogram. Using Pythagorean theorem on
,
. Therefore, base
. Thus the area of the parallelogram is the base times the height, which is
and the answer is
~KingRavi
Let the circle tangent to at
separately, denote that
Using POP, it is very clear that , let
, using LOC in
,
, similarly, use LOC in
, getting that
. We use the second equation to minus the first equation, getting that
, we can get
.
Now applying LOC in , getting
, solving this equation to get
, then
,
, the area is
leads to
~bluesoul
Denote by the center of the circle. Denote by
the radius of the circle. Denote by
,
,
the points that the circle meets
,
,
at, respectively.
Because the circle is tangent to ,
,
,
,
,
,
.
Because ,
,
,
are collinear.
Following from the power of a point, . Hence,
.
Following from the power of a point, . Hence,
.
Denote . Because
and
are tangents to the circle,
.
Because is a right trapezoid,
. Hence,
. This can be simplified as [ 6 x = r^2 . hspace{1cm} (1) ]
In , by applying the law of cosines, we have begin{align*} AC^2 & = AB^2 + CB^2 - 2 AB cdot CB cos B \ & = AB^2 + CB^2 + 2 AB cdot CB cos A \ & = AB^2 + CB^2 + 2 AB cdot CB cdot frac{AE - BF}{AB} \ & = AB^2 + CB^2 + 2 CB left( AE - BF right) \ & = left( 6 + x right)^2 + left( 20 + x right)^2 + 2 left( 20 + x right) left( 6 - x right) \ & = 24 x + 676 . end{align*}
Because , we get
. Plugging this into Equation (1), we get
.
Therefore, begin{align*} {rm Area} ABCD & = CB cdot EF \ & = left( 20 + x right) cdot 2r \ & = 147 sqrt{3} . end{align*}
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Let be the circle, let
be the radius of
, and let the points at which
is tangent to
,
, and
be
,
, and
, respectively. Note that PoP on
and
with respect to
yields
and
. We can compute the area of
in two ways:
1. By the half-base-height formula, .
2. We can drop altitudes from the center of
to
,
, and
, which have lengths
,
, and
. Thus,
.
Equating the two expressions for and solving for
yields
.
Let . By the Parallelogram Law,
. Solving for
yields
. Thus,
, for a final answer of
.
~ Leo.Euler
For any finite set , let
denote the number of elements in
. Define
where the sum is taken over all ordered pairs
such that
and
are subsets of
with
. For example,
because the sum is taken over the pairs of subsets
giving
. Let
, where
and
are relatively prime positive integers. Find the remainder when
is divided by 1000.
Let's try out for small values of to get a feel for the problem. When
is obviously
. The problem states that for
is
. Let's try it out for
.
Let's perform casework on the number of elements in .
In this case, the only possible equivalencies will be if they are the exact same element, which happens times.
In this case, if they share both elements, which happens times, we will get
for each time, and if they share only one element, which also happens
times, we will get
for each time, for a total of
for this case.
In this case, the only possible scenario is that they both are the set , and we have
for this case.
In total, .
Now notice, the number of intersections by each element , or in general,
is equal for each element because of symmetry - each element when
adds
to the answer. Notice that
- let's prove that
(note that you can assume this and answer the problem if you're running short on time in the real test).
Let's analyze the element - to find a general solution, we must count the number of these subsets that
appears in. For
to be in both
and
, we need
and
(Basically, both sets contain
and another subset of
through
not including
).
For any that is the size of both
and
, the number of ways to choose the subsets
and
is
for both subsets, so the total number of ways to choose the subsets are
. Now we sum this over all possible
's to find the total number of ways to form sets
and
that contain
. This is equal to
. This is a simplification of Vandermonde's identity, which states that
. Here,
,
and
are all
, so this sum is equal to
. Finally, since we are iterating over all
's for
values of
, we have
, proving our claim.
We now plug in to the expression we want to find. This turns out to be
. Expanding produces
.
After cancellation, we have
and
don't have any common factors with
, so we're done with the simplification. We want to find
~KingRavi
We take cases based on the number of values in each of the subsets in the pair. Suppose we have elements in each of the subsets in a pair (for a total of n elements in the set). The expected number of elements in any random pair will be
by linearity of expectation because for each of the
elements, there is a
probability that the element will be chosen. To find the sum over all such values, we multiply this quantity by
. Summing, we get
Notice that we can rewrite this as
We can simplify this using Vandermonde's identity to get
. Evaluating this for
and
gives
Evaluating the numerators and denominators mod
gives
- pi_is_3.14
For each element , denote
, where
(resp.
).
Denote .
Denote .
Hence,
Therefore,
This is in the lowest term. Therefore, modulo 1000,
~Steven Chen (www.professorchenedu.com
Let be the set of all rational numbers that can be expressed as a repeating decimal in the form
where at least one of the digits
or
is nonzero. Let
be the number of distinct numerators obtained when numbers in
are written as fractions in lowest terms. For example, both
and
are counted among the distinct numerators for numbers in
because
and
Find the remainder when
is divided by
,
.
Then we need to find the number of positive integers less than 10000 can meet the requirement.Suppose the number is x.
Case 1: (9999, x)=1. Clearly x satisfies.
Case 2: 3|x but x is not a multiple of 11 or 101. Then the least value of abcd is 9x, so that , 334 values from 3 to 1110.
Case 3: 11|x but x is not a multiple of 3 or 101. Then the least value of abcd is 11x, so that , 55 values from 11 to 902.
Case 4: 101|x. None.
Case 5: 3, 11|x. Then the least value of abcd is 11x, 3 values from 33 to 99.
To sum up, the answer is
Let
and
be positive real numbers satisfying the system of equations:
Then
can be written as
where
and
are relatively prime positive integers. Find
First, we note that we can let a triangle exist with side lengths ,
, and opposite altitude
. This shows that the third side, which is the nasty square-rooted sum, is going to have the length equal to the sum on the right - let this be
for symmetry purposes. So, we note that if the angle opposite the side with length
has a value of
, then the altitude has length
and thus
so
and the triangle side with length
is equal to
.
We can symmetrically apply this to the two other triangles, and since by law of sines, we have is the circumradius of that triangle. Hence. we calculate that with
, and
, the angles from the third side with respect to the circumcenter are
, and
. This means that by half angle arcs, we see that we have in some order,
,
, and
(not necessarily this order, but here it does not matter due to symmetry), satisfying that
,
, and
. Solving, we get
,
, and
.
We notice that
- kevinmathz
(This eventually whittles down to the same concept as Solution 1)
Note that in each equation in this system, it is possible to factor ,
, or
from each term (on the left sides), since each of
,
, and
are positive real numbers. After factoring out accordingly from each terms one of
,
, or
, the system should look like this:
This should give off tons of trigonometry vibes. To make the connection clear,
,
, and
is a helpful substitution:
From each equation
can be factored out, and when every equation is divided by 2, we get:
which simplifies to (using the Pythagorean identity
):
which further simplifies to (using sine addition formula
):
Without loss of generality, taking the inverse sine of each equation yields a simple system:
giving solutions
,
,
. Since these unknowns are directly related to our original unknowns, there are consequent solutions for those:
,
, and
. When plugging into the expression
, noting that
helps to simplify this expression into:
Now, all the cosines in here are fairly standard: ,
,
and
. With some final calculations:
This is our answer in simplest form
, so
-Oxymoronic15
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