AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, in square units, is![\[ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 46 \qquad
\textbf{(D)}\ 66 \qquad
\textbf{(E)}\ 74
\]](http://latex.artofproblemsolving.com/8/2/f/82ffc15072ca99f6cd1abc312781eb4827817356.png)
is rotated 
degrees about one side. The space through which the hexagon travels during the rotation forms a solid. Find the volume of this solid. is rotated degrees about one side. The space through which the octagon travels through during the rotation forms a solid. Find the volume of this solid.
be an acute scalene triangle. Let 
and 
be two distinct interior points of the segment 
such that 
. Suppose that:
and 
are the feet of the perpendiculars from from 
to the lines 
and 
respectively.
and 
are the feet of the perpendiculars from 
to the lines and respectively.
and 
intersect on the line .
[hide=P#]a1b2c3[/hide] [hide=S#]a1b2c3[/hide]
be a triangle with incenter 
, and let 
be a point on side . Points and are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side such that and are the angle bisectors of angles 
and 
respectively. Let 
be the circle tangent to segment 
, the extension of 
past , and the extension of 
past . Prove that is tangent to the circumcircle of triangle .
of the triangle 
which doesn't coincide with 
or 
intersectes the altitudes from 
and 
at 
and 
respectively. Let 
be the reflection of in and 
be the reflection of in 
Prove that the circles 
and 
are tangent.
be positive real numbers such that 
Prove that we have![\[(a+b)(b+c)(c+a)\geq 8.\]](http://latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
, points 
are constructed as follows: Points 
and 
lie on the extension of side 
such that 
. Points 
and 
lie on the extension of side 
such that 
. Points 
and 
lie on the extension of side 
such that 
..![\[
\angle ACE + \angle FBD + \angle EAC = 180^\circ.
\]](http://latex.artofproblemsolving.com/c/b/8/cb88e7d6ec129d50bde13e6a08ee16b275f42042.png)
lies inside of scalene triangle with incenter such that 
Lines 
and 
intersect line 
respectively at 
and 
. Line through parallel to 
intersects at and line through parallel to 
intersects at . Prove that triangles 
and are similar.
is the -excenter of the triangle and is the circumcircle of this triangle. 
is the middle of arc of which doesn't contain 
. 
meets at 
. Prove that
, point 
is marked. Point 
is on segment 
, and point 
is on segment 
such that the circumcircles of triangles 
and 
are tangent to line 
. Lines are drawn through points 
and 
passing through the center of the circumcircle of triangle 
, and through points and passing through the center of the circumcircle of triangle 
. Prove that there exists a circle tangent to all four drawn lines.
Then there are two cases either 
or 
.When we are admitting the case we cannot divide by 
. So, in the case we apply divison by 
then is a solid prerequisite to do so.Thus, 
from 
we must take the assumption in hand that . For example take the extension of the same problem in 
then the same problem reads 
,implying only one solution with optimistic repetation of root , two times that is the multiplicity of in 
. Thankfully, we are lucky enough that we are in the field of 
.The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of .
Then there are two cases either or .When we are admitting the case we cannot divide by . So, in the case we apply divison by then is a solid prerequisite to do so.Thus, from we must take the assumption in hand that . For example take the extension of the same problem in then the same problem reads ,implying only one solution with optimistic repetation of root , two times that is the multiplicity of in . Thankfully, we are lucky enough that we are in the field of .The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of ..
Then there are two cases either or .When we are admitting the case we cannot divide by . So, in the case we apply divison by then is a solid prerequisite to do so.Thus, from we must take the assumption in hand that . For example take the extension of the same problem in then the same problem reads ,implying only one solution with optimistic repetation of root , two times that is the multiplicity of in . Thankfully, we are lucky enough that we are in the field of .The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of ...The assumption whenever we get 
else we get the case .I can explain you more but the fact is I cannot use argument of field theory to explain it in total details.The reason why it's actually the case lies in field theory logics.
Which means that the roots of this equation have to be real so we can use an method that is lost in the darkness called factoring. We can factor out the from each of the terms on the left hand side and get which with more logic we can find that the possible outcomes is that if the Parantheses are 0 or the x=0 so first we can subsitute a value of x into that to make the value 0 so we get that x=5 and we finally get the solutions of and and to wrap up our proof we can prove that factoring is the best way to go because with quadratics you would only find 2 possibiliteis or 1 depending on the plus minus. And the other thing is that if you divide by x and just solve it with algebra then you will only get the solution of 5. Thus, proving that factoring is the best method out of all of them. We can use the remainder theorem to prove this which can be done easily. 
is lost? right. Well forget about field theory and all that, Let me explain it in layman's term what is actually happening. In second solution , the solution is not actually lost. The reason we are getting but not is beacuse when we are dividing by we making an assumption that and since we are making this assumption the solution is lost. For example when we divide by 
the solution is lost why 
and we are assuming 
which is equivalent to 
. Now divison by zero is not possible which is not at all very easy to explain. Now and is not possible at the same time. Thus either or . and 
can be same when we are in a field of 
. If you are avoid knowing what is field that's perfectly fine to learn later, but just in layman's term note that 
is possible in finite fields of charecteristic .
and height 
:
Volume of cone frustum with base radii 
,
:o
so 
and since 
is just 
therefore 
so 
