and 
such that 
and 
.
if within that number exists a digit 
that appears exactly 
times in that number, hence find the number of 
and 
such that:
be a triangle, 
is incenter of triangle 
perpendicular to 
at 
and 
perpendicular to 
at 
,
.
be the solid square region with vertices at![\[
(0, 0), \left(\frac{1}{2}, \frac{1}{2}\right), (1, 0), \left(\frac{1}{2}, -\frac{1}{2}\right).
\]](http://latex.artofproblemsolving.com/a/9/8/a9898f9237f07eaef5d17a9ac689d80ca631863f.png)
consists of two copies of 
:
counterclockwise around the origin and scaled by a factor of 
,
.
below.
be the limiting region of the sequence 
.
for relatively prime positive integers 
and 
.
.
shows that after each species divides and then each amoeba cell preys on exactly one bacterial cell, it turns out kill bacteria. Determine the ratio of the number of amoeba to the number of bacteria on the first day.
is a positive integer. Let 
.
cm. The first nine balls are placed on the table so that
cm, 
cm, and 
cm. Find the maximum possible area of the rectangle can be made in the triangle 
people waiting in line to buy tickets to a show with the price of one ticket is 
Rp.. Known 
Rp. in banknotes and the rest is only has a banknote of 
with
regions. Every pair of lines in the same group are parallel. Let 
and 
respectively be the number of lines in groups 1, 2, and 3. If no lines in group 3 go through the intersection of any two lines (in groups 1 and 2, of course), then the least number of lines required in order to have more than 2018 regions is ....
where 
are squares with both planes being parallel. The length of the sides of 
and 
,
.
is perpendicular to the plane 
and 
and calculate the volume of the 3D figure which is the intersection of pyramids 
is on the 3rd row, 4th column, and on the 6th diagonal. Meanwhile the position of the number 
is on the 3rd row, 5th column, and 7th diagonal.
based on its row, column, and diagonal.
is the set of 3-digit integers not containing the digit 
.
and 
dan 
.
is a fraction in its simplest form.
.
be given in the following graphs.
or 
,
,
,
,
,
.
.
with 
for all 
where 
is the domain of 
.
cm and 
cm. All four of its vertices lie on a circle. Calculate the area of quadrilateral 
,
and 
.
,
(the original problem does not have an overline, which I fixed) is arranged from the set 
.
.
(i.e., the square on 
has 
,
,
to the centers of the opposite squares are concurrent.
.
?
?
and 
are points on sides 
and 
.
and 
,
side length of 
area of 
is a right angle, and 
,
.
is a right angle, 
,
.
and 
both contain a right angle and 
through AA postulate. Because they are similar, 
,
,
,
.
.
,
.
Y by
My question is after the problem and my solution.
Problem:
In the quadrilateral
, 
, 
, 
.
(1) Find
.
(2) If
, find 
. (2018 China Gaokao Syllabus I #17)
Solution:
Draw a diagram (see the attachment):
(1) First,
. Since 
, we can plug it back and find that 
. Therefore, through the Pythagorean Identities, we find that 
.
(2) Notice that since, its sine is 
. Therefore, using the fact that 
, we can say: 
.
Expand using sum of sines:
.
Using
, call 
equal to 
(notice that since 
is acute, its sine must be positive, so that's why I used the cosine to set the value to avoid problems), 
. Plug it back in and solve the quadratic equation, we find that 
. Now using the law of cosines, we find that 
.
Wow! Such a long bash! Thanks for reading the solution. Now, here's the question:
In the answer key to part (2) of this problem, it said:
Because of the problem's given information and part (1) [its solution is identical to mine, just differently worded], it is obvious that
.
How come? I bashed all of this out to obtain that fact. How come it's "obvious?" Please explain. Notice that if the answer key omitted these details, it wouldn't be an answer key because the bashing is vital to this problem it looks like!
Thanks!
Problem:
In the quadrilateral
, 
, 
, 
.(1) Find
.(2) If
, find 
. (2018 China Gaokao Syllabus I #17)Solution:
Draw a diagram (see the attachment):
(1) First,
. Since 
, we can plug it back and find that 
. Therefore, through the Pythagorean Identities, we find that 
.(2) Notice that since
, its sine is 
. Therefore, using the fact that 
, we can say: 
.Expand using sum of sines:
.Using
, call 
equal to 
(notice that since 
is acute, its sine must be positive, so that's why I used the cosine to set the value to avoid problems), 
. Plug it back in and solve the quadratic equation, we find that 
. Now using the law of cosines, we find that 
.Wow! Such a long bash! Thanks for reading the solution. Now, here's the question:
In the answer key to part (2) of this problem, it said:
Because of the problem's given information and part (1) [its solution is identical to mine, just differently worded], it is obvious that
. How come? I bashed all of this out to obtain that fact. How come it's "obvious?" Please explain. Notice that if the answer key omitted these details, it wouldn't be an answer key because the bashing is vital to this problem it looks like!
Thanks!
Attachments:
So the fact that 
is a direct result from the fact that 
The proof of this isn't too hard. Just consider the unit circle. Notice that rotating the point 
by an angle of 
counterclockwise takes it to the point 
and rotating it 
Also notice that these two points are symmetric about 
so the points 
and 
are the same, so 
and 
