[/quote]
,
![\[\left(-\cos(1^{\circ})\right)\left(-\cos(2^{\circ})\right)\left(-\cos(4^{\circ})\right) \cdots \left(-\cos{(256^{\circ})}\right)\]](http://latex.artofproblemsolving.com/4/4/e/44e5bbdf821aed7bb2d855efcf39af763ebaea48.png)
on a blackboard. Then, he erases each term on the blackboard with probability 
.
with 
.
takes place for all choiches of the 
signs. Show that there exists a line 
through 
such that all points 
are all on one side of 
.
is the midpoint of 
is the foot of altitude from point 
.
and 
is equilateral, and 
and this points lie in the same half-plane with respect to 
and 
are tangent.
and 
are given in space, such that 
.
and 
and 
.
is a right triangle;
and 
are perpendicular.
,
on 
,
be points on lines 
,
,
concur at a point 
be a point on line 
,
,
be a point on line 
be a point on line 
,
,
concur at a single point.
and 
and one of its tangents be 
.
is the foot of the altitude from 
points 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles 
.
.
is given.
is the interior and the incircle of 
at 
.
be a tangent from D to the circumcircle of 
,
and 
on 
meet at one point.
and 
are drawn such that 
and 
respectively. It is known that 
.
,
,
centimeter. Find the length of diagonal 
be the circle through 
,
and 
.
be the other intersection point of 
be the reflection of 
.
,
.
passes through the midpoint of segment 
and 
,and 
.
intersects 
intersects 
)
.
Y by evt917, pingpongmerrily, valenbb, clarkculus, alexanderhamilton124, At777, GodGodGodGodGoose, Erratum
There are math questions at the end of the post.
Celebrating 22nd December
And as a result I am quitting AoPS (atleast as of now)
Why?
I will probably update my blog(?).
So yeah this concludes this post(?). I probably have not written whatever I wanted to but AoPS has been basically the website I have been the most active on for the past year so leaving this hurts. (Not like I am going for forever lol)
Anyways as promised here are a "few" (not so original) questions.
Questions
Celebrating 22nd December
Today is the birthday of our beloved Srinivasa Ramanujan Aiyangar and also India's National Mathematics Day! (not a coincidence!). But coincidently, it is my AoPS account's birthday too! (I swear it was a coincidence!). My account has turned exactly 
year today and has 
posts as of now (excluding this) and has received 
upvotes.
year today and has 
posts as of now (excluding this) and has received 
upvotes.And as a result I am quitting AoPS (atleast as of now)
Why?
I have a big exam upcoming in the next 
months and it is probably the biggest exam I have ever given. So yeah I need to prepare for that and I think I should not spend much time on AoPS anymore as of now. I will probably join AoPS after 
months. I know no one asked but just in case you guys may be surprised as to why are there no more billion HSM posts per day.
months and it is probably the biggest exam I have ever given. So yeah I need to prepare for that and I think I should not spend much time on AoPS anymore as of now. I will probably join AoPS after 
months. I know no one asked but just in case you guys may be surprised as to why are there no more billion HSM posts per day.I will probably update my blog(?).
So yeah this concludes this post(?). I probably have not written whatever I wanted to but AoPS has been basically the website I have been the most active on for the past year so leaving this hurts. (Not like I am going for forever lol)
Anyways as promised here are a "few" (not so original) questions.
Questions
Geometry:
Here is a collection of geometry problems and solutions for new ideas:
Q1
S1![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
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label("$c$", (-5.36,1.44), NE * labelscalefactor,zzttqq);
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label("$b$", (1.9,1.4), NE * labelscalefactor,zzttqq);
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dot((-0.08,-1.48),linewidth(4pt) + dotstyle);
label("$E$", (0,-1.32), NE * labelscalefactor);
label("$g$", (-1.96,1.02), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](//latex.artofproblemsolving.com/2/7/4/274d0dbc781e3bc128d6477a45f887b2ed13f17b.png)
. 
as 
.
Using Internal Angle Bisector Theorem in
.
, or, 
.
Using Pythagorean Theorem in,
, or, 
.
Using Pythagorean Theorem in
,
.
Thus
.
Thus the closest integer is
.
S1 by lbh_qys
Q2
S2
Q3
S3
Q4
S4![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
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pen dotstyle = black; /* point style */
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draw((-3.64,-3.22)--(1.84,-3.28), linewidth(2) + zzttqq);
draw((1.84,-3.28)--(1.9,2.2), linewidth(2) + zzttqq);
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draw((-3.5946697742558436,0.9201606179663049)--(1.8594891123393478,-1.499994406339482), linewidth(2));
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draw((xmin, 91.33333333333121*xmin + 148.3333333333298)--(xmax, 91.33333333333121*xmax + 148.3333333333298), linewidth(2)); /* line */
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dot((-3.64,-3.22),dotstyle);
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dot((1.84,-3.28),dotstyle);
label("$C$", (1.92,-3.08), NE * labelscalefactor);
dot((1.9,2.2),dotstyle);
label("$D$", (1.98,2.4), NE * labelscalefactor);
dot((-3.5946697742558436,0.9201606179663049),dotstyle);
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dot((1.8594891123393478,-1.499994406339482),dotstyle);
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label("$j$", (-1,-0.58), NE * labelscalefactor);
dot((-1.6595804754611438,-3.2416834254511557),dotstyle);
label("$G$", (-1.58,-3.04), NE * labelscalefactor);
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dot((0.760574548844643,2.2124754611440354),linewidth(4pt) + dotstyle);
label("$H$", (0.84,2.38), NE * labelscalefactor);
label("$l$", (-0.76,-0.38), NE * labelscalefactor);
label("$m$", (-8.76,-1.22), NE * labelscalefactor);
label("$n$", (-1.42,4.92), NE * labelscalefactor);
dot((-1.5995804754611425,2.238316574548844),linewidth(4pt) + dotstyle);
label("$I$", (-1.52,2.4), NE * labelscalefactor);
dot((-3.620510887660651,-1.4399944063394812),linewidth(4pt) + dotstyle);
label("$J$", (-3.54,-1.28), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](//latex.artofproblemsolving.com/c/8/c/c8c25eae8a34b902f4eb033642872f74f9afe087.png)
We rename the points as per the diagram.
Let
and 
.
Then,
as 
, and 
and 
.
Thus the required length is
.
S4 by Mathzeus1024
Q5
S5
Q6
S6
Combinatorics:
Here is a collection of combinatorics problems and solutions for new ideas:
P1
S1 (Gap Method)
Q2
S2
Comment
Functional Equations:
Here are a few problems of Functional Equations:
Q1
S1
Q2
S2
Q3
S3
Here is a collection of geometry problems and solutions for new ideas:
Q1
In a triangle 
, the altitude 
and the median 
divide 
into three equal parts. If 
, then the nearest integer to 
is?
, the altitude 
and the median 
divide 
into three equal parts. If 
, then the nearest integer to 
is?S1
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
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draw((-3.26,3.82)--(-6.92,-1.34), linewidth(2) + zzttqq);
draw((-6.92,-1.34)--(6.76,-1.62), linewidth(2) + zzttqq);
draw((6.76,-1.62)--(-3.26,3.82), linewidth(2) + zzttqq);
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label("$A$", (-3.18,4.02), NE * labelscalefactor);
dot((-6.92,-1.34),dotstyle);
label("$B$", (-6.84,-1.14), NE * labelscalefactor);
dot((6.76,-1.62),dotstyle);
label("$C$", (6.84,-1.42), NE * labelscalefactor);
label("$c$", (-5.36,1.44), NE * labelscalefactor,zzttqq);
label("$a$", (-0.1,-1.78), NE * labelscalefactor,zzttqq);
label("$b$", (1.9,1.4), NE * labelscalefactor,zzttqq);
label("$f$", (-3.08,4.92), NE * labelscalefactor);
dot((-3.36710245870117,-1.4127201251143038),linewidth(4pt) + dotstyle);
label("$D$", (-3.28,-1.26), NE * labelscalefactor);
dot((-0.08,-1.48),linewidth(4pt) + dotstyle);
label("$E$", (0,-1.32), NE * labelscalefactor);
label("$g$", (-1.96,1.02), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/2/7/4/274d0dbc781e3bc128d6477a45f887b2ed13f17b.png)
. 
as 
.Using Internal Angle Bisector Theorem in
.
, or, 
.Using Pythagorean Theorem in
,
, or, 
.Using Pythagorean Theorem in
,
.Thus
.Thus the closest integer is
.S1 by lbh_qys
Let 
be the circumcenter, then 
, which indicates that 
, , and 
are colinear. Therefore, 
or 
. However, when , 
and 
coincide, and the angle is not trisected. Thus, , and triangle 
is a right-angled triangle. Consequently, the altitude divides the 
angle into 
and 
, which implies the triangle is a right-angled triangle with angles , , and . Therefore, 
, and the closest integer is 
.
be the circumcenter, then 
, which indicates that 
, , and 
are colinear. Therefore, 
or 
. However, when , 
and 
coincide, and the angle is not trisected. Thus, , and triangle 
is a right-angled triangle. Consequently, the altitude divides the 
angle into 
and 
, which implies the triangle is a right-angled triangle with angles , , and . Therefore, 
, and the closest integer is 
.Q2
The sides 
of a triangle satisfy the relations 
and 
. Then the measure of 
, in degrees, is?
of a triangle satisfy the relations 
and 
. Then the measure of 
, in degrees, is?S2
Q3
Let 
be a triangle in which 
. From the vertex 
, draw the altitude 
to meet 
at 
. Assume that 
and 
. Then 
equals?
be a triangle in which 
. From the vertex 
, draw the altitude 
to meet 
at 
. Assume that 
and 
. Then 
equals?S3
Let 
. 
(Pythagorean Theorem). Thus 
and 
(Pythagorean Theorem).
. 
(Pythagorean Theorem). Thus 
and 
(Pythagorean Theorem).Q4
Let 
be a square of side length . Let 
be points in the interiors of the sides 
, respectively, such that and intersect at right angles. If 
then find the length of .
be a square of side length . Let 
be points in the interiors of the sides 
, respectively, such that and intersect at right angles. If 
then find the length of .S4
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -8.92, xmax = 8.92, ymin = -5.4, ymax = 5.4; /* image dimensions */
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draw((-3.58,2.26)--(-3.64,-3.22)--(1.84,-3.28)--(1.9,2.2)--cycle, linewidth(2) + zzttqq);
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dot((1.84,-3.28),dotstyle);
label("$C$", (1.92,-3.08), NE * labelscalefactor);
dot((1.9,2.2),dotstyle);
label("$D$", (1.98,2.4), NE * labelscalefactor);
dot((-3.5946697742558436,0.9201606179663049),dotstyle);
label("$E$", (-3.52,1.12), NE * labelscalefactor);
dot((1.8594891123393478,-1.499994406339482),dotstyle);
label("$F$", (1.94,-1.3), NE * labelscalefactor);
label("$j$", (-1,-0.58), NE * labelscalefactor);
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label("$G$", (-1.58,-3.04), NE * labelscalefactor);
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label("$l$", (-0.76,-0.38), NE * labelscalefactor);
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label("$n$", (-1.42,4.92), NE * labelscalefactor);
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[/asy]](http://latex.artofproblemsolving.com/c/8/c/c8c25eae8a34b902f4eb033642872f74f9afe087.png)
We rename the points as per the diagram.
Let
and 
.Then,
as 
, and 
and 
.Thus the required length is
.S4 by Mathzeus1024
Let the points 
be located in the 
plane. If 
, then:
(i).
If
, then 
(ii).
Finally,
.
be located in the 
plane. If 
, then:
(i).If
, then 
(ii).Finally,
.Q5
Let 
be a point inside a triangle with 
. Let 
and 
be the images of under reflection in 
and 
respectively. The distance between the circumcenters of triangles and 
is?
be a point inside a triangle with 
. Let 
and 
be the images of under reflection in 
and 
respectively. The distance between the circumcenters of triangles and 
is?
S5
Let 
and 
and 
. Assume 
then 
and 
. Observe that the circumcenter of 
is just 
. Observe that the circumcenter of 
is just 
. Thus the distance is just the circumradius or 
.
and 
and 
. Assume 
then 
and 
. Observe that the circumcenter of 
is just 
. Observe that the circumcenter of 
is just 
. Thus the distance is just the circumradius or 
.Q6
Three circles of radii 
and 
units respectively touch each other externally in the place. The circumradius of the triangle formed by joining the centers of the circles is?
and 
units respectively touch each other externally in the place. The circumradius of the triangle formed by joining the centers of the circles is?S6
Observe the triangle formed is a 
right angled triangle thus the circumradius is 
.
right angled triangle thus the circumradius is 
.Combinatorics:
Here is a collection of combinatorics problems and solutions for new ideas:
P1
Connie finds a whiteboard that has magnet letters spelling MISSISSIPPI on it. She can rearrange the letters, in which identical letters are indistinguishable. If she uses all the letters and does not want to place any Is next to each other, how many distinct rearrangements are possible?
S1 (Gap Method)
We first arrange the M,P,S and later put the Is. Ways to arrange the M,P,S 
. We have 
places to choose form and way to arrange the Is in it. Thus the final answer is 
.
. We have 
places to choose form and way to arrange the Is in it. Thus the final answer is 
.Q2
Carson is planning a trip for 
people. Let 
be the number of cars that will be used and 
be the number of people per car. What is the smallest value of such that there are exactly possibilities for and so that is an integer, 
, and exactly one person is left without a car?
people. Let 
be the number of cars that will be used and 
be the number of people per car. What is the smallest value of such that there are exactly possibilities for and so that is an integer, 
, and exactly one person is left without a car?S2
Ofcourse 
. Now 
. As there are possible values for (and the corresponding values of ) thus the total number of factors of 
is 
. can be of the form 
or 
. We bash to find the least is possible in the second case with 
and 
thus resulting the final in being 
.
. Now 
. As there are possible values for (and the corresponding values of ) thus the total number of factors of 
is 
. can be of the form 
or 
. We bash to find the least is possible in the second case with 
and 
thus resulting the final in being 
.Comment
Observing 
was quite natural. Now the main thing is observing that has factors. This gives us an idea that whenever in case of product and stuff with a certain quantity given, we should check for factors and stuff.
was quite natural. Now the main thing is observing that has factors. This gives us an idea that whenever in case of product and stuff with a certain quantity given, we should check for factors and stuff.Functional Equations:
Here are a few problems of Functional Equations:
Q1
Let 
satisfy the equation 
. Then :
i) Prove that
is bijective.
ii) Prove that is multiplicative
satisfy the equation 
. Then :i) Prove that
is bijective.ii) Prove that
is multiplicativeS1
Claim: is injective.
Proof: Say is not injective. Then 
such that 
.
But a function can't give two distinct values, unless 
this is not true so must be injective.
Claim: is surjective.
Proof: Fix
, or else .
To obtain any 
just let 
.
Now,
Thus is multiplicative.
is injective.Proof: Say
is not injective. Then 
such that 
.
But a function can't give two distinct values, unless 
this is not true so must be injective.Claim:
is surjective.Proof: Fix
, or else .
To obtain any 
just let 
.Now,
Thus is multiplicative.Q2
Find all functions 
which satisfy for all :
for 
which satisfy for all :
for 
S2
Note that
![\[f(x)=x^2\cdot f\left(\frac{1}{x}\right)=x^2\cdot f\left(1+\frac{1-x}{x}\right)=x^2+x^2\cdot f\left(\frac{1-x}{x}\right)=x^2+x^2\cdot \frac{(1-x)^2}{x^2}\cdot f\left(\frac{x}{1-x}\right)=\]](//latex.artofproblemsolving.com/4/d/a/4da809b44b08f413bde69c4df27874c14bc7e4b8.png)
![\[x^2+(1-x)^2\cdot f\left(\frac{1}{1-x}-1\right)=x^2+(1-x)^2\cdot f\left(\frac{1}{1-x}\right)-(1-x)^2=x^2-(1-x)^2+f(1-x)=2x+f(-x)=2x-f(x)\]](//latex.artofproblemsolving.com/3/d/d/3ddf8024f0a68b7279bee29f51e58e913fd1d3aa.png)
and hence 
holds for all 
which easily extends to for all which is indeed a solution.
![\[f(x)=x^2\cdot f\left(\frac{1}{x}\right)=x^2\cdot f\left(1+\frac{1-x}{x}\right)=x^2+x^2\cdot f\left(\frac{1-x}{x}\right)=x^2+x^2\cdot \frac{(1-x)^2}{x^2}\cdot f\left(\frac{x}{1-x}\right)=\]](http://latex.artofproblemsolving.com/4/d/a/4da809b44b08f413bde69c4df27874c14bc7e4b8.png)
![\[x^2+(1-x)^2\cdot f\left(\frac{1}{1-x}-1\right)=x^2+(1-x)^2\cdot f\left(\frac{1}{1-x}\right)-(1-x)^2=x^2-(1-x)^2+f(1-x)=2x+f(-x)=2x-f(x)\]](http://latex.artofproblemsolving.com/3/d/d/3ddf8024f0a68b7279bee29f51e58e913fd1d3aa.png)
and hence 
holds for all 
which easily extends to for all which is indeed a solution.Q3
Find all functions 
such that
such that
S3
The only constant solutions are 
and 
.
So let us from now look only for non constant solutions.
Let
be the assertion 
Let
and so 
and 
since 
is non constant;
and so :
and so 
implies 
which gives immediately 
, 
and 
which indeed is a solution.
implies 
which gives immediately 
and 
and 
which indeed is a solution.
implies 
which gives immediately 
which indeed is a solution.
Hence the five solutions :
, and
and and
and 
.So let us from now look only for non constant solutions.
Let
be the assertion 
Let
and so 
and 
since 
is non constant;
and so :
and so 
implies 
which gives immediately 
, 
and 
which indeed is a solution.
implies 
which gives immediately 
and 
and 
which indeed is a solution.
implies 
which gives immediately 
which indeed is a solution.Hence the five solutions :
, and and and 
