AoPS Academy
from the non-zero reals to the non-zero reals, such that ![\[f(x+y) \left(f(x) + f(y)\right) = f(xy)\]](http://latex.artofproblemsolving.com/2/f/e/2fe8a09e628f60faedaa07ceb154891375763a83.png)
for all non-zero reals 
such that 
.
. Prove that
Let 
. Prove that
Let 
. Prove that
+1 w
be a real numbers such that 
Prove that
Where 
so that both and 
have odd numbers of divisors.
which is defined as all the natural numbers listed in order, side by side, behind a decimal point, without commas? If so, what is it?
and 
are primes, then can 
be a perfect square?
be a trapezoid such that 
lie on a circle with center 
, and side 
is parallel to side 
. Diagonals 
and 
intersect at point 
, and 
. It is given that 
. It is also known that the difference in length between and can be expressed in the form 
, where 
and 
are positive integers and is square-free. Compute the value of 
.
, where 
always occurs at 
, above the xy-plane, such that the pyramid 
is a regular tetrahedron.
be a trapezium in which 
and 
. Let 
be then midpoint of the diagonal 
. If ![$[ABCD] = n \times [CDE]$](http://latex.artofproblemsolving.com/e/9/3/e9347ba2c8e9396adfe716159e711408a3b9dc69.png)
, what is the value of ?![$[t]$](http://latex.artofproblemsolving.com/8/a/8/8a86702dc60b9db916798b1115a6b6822bf828b9.png)
denotes the area of the geometrical figure
.)
be an integer. Let 
, 
, ..., 
be positive real numbers such that ![\[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\]](http://latex.artofproblemsolving.com/1/a/b/1ab410ebc17d9f6c38a3e7ae0f6efa955ffddfd3.png)
Show that 
, 
, 
are side lengths of a triangle for all 
, 
, 
with 
.
. it is easy to see that the right is strictly increasing in 
. so if we show that the reverse inequality holds for 
, we are done. a quick calculation is enough to show that(it is too boring to post). but if we know is true for n-1, then let's do the following computation:
,
be an integer. Suppose the inequality
. Prove that 
are the three sides of some triangle for all 
.
terms is at least 
, the whole sum will be at least 
and the inequality will be violated.
. Then 
when 
, we will be done by the argument above.
, so 
and 
verify 
then any three of them are sides of a triangle. I think I've managed to solve this problem and if my solution is correct, then it's a really hard and nice problem.
![\[ (t_1+ \cdots + t_n) ( \frac{1}{t_1} + \cdots + \frac{1}{t_n}) = n^2 + \sum_{i<j} (\sqrt{\frac{t_i}{t_j}} - \sqrt{\frac{t_j}{t_i}} )^2 \]](http://latex.artofproblemsolving.com/8/2/e/82e1a00585a2b5adbd6c59119f4bf55c41b04f5e.png)
then![\[{{ (\sqrt{a/b} - \sqrt{b/a})^2 + (\sqrt{a/c} - \sqrt{c/a})^2 =
\frac{ (a-b)^2}{ab} } + \frac{ (a-c)^2}{ac} }
\geq \frac{(a-b)c}{ab} + \frac{(a-c)b}{ac}
= c/b + b/c -(b+c)/a \geq 2 - 1 = 1 \]](http://latex.artofproblemsolving.com/3/c/1/3c1c7545ca6ebf41b70af49f67869215ec2039d5.png)
(t_1+t_2+...+t_n)(1/t_1+1/t_2+...+1/t_n)=(3\times(t_1+t_2+t_3)/3 +t_4+...+t_n)(3\times(1/t_1+1/t_2+1/t_3)/3 +1/t_4+...+1/t_n)\geq(3\times \sqrt((t_1+t_2+t_3)/3\times(1/t_1+1/t_2+1/t_3)/3) +1+...+1)^2
.
numebr on his list? Write its last two digits as your answer.
grid of square tiles. Robert chooses four of these squares uniformly at random. The probability that the centers of these four squares form a square themselves is 
where 
are coprime naturals. Write the largest 
digit odd factor of 
.
.
which touches one face of the cube and passes through all 
vertices of the face opposite to it. Find the integer closest to 
are distinct numbers in AP such that 
are also in AP. Find the smallest possible value of 
(in degrees).
that satisfy 
.n^2+1>(\sqrt((t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)) +n-3)^2(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)<(\sqrt(n^2+1) -(n-3))^2n\geq3, it follows(\sqrt(n^2+1) -(n-3))^2<10 <=> 2n^2 -6n<(2n-6) \sqrt(n^2+1)
(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)<10
t_1 \geq t_2 \geq t_3. Then
t_1+t_2 \geq t_3, t_1+t_3 \geq t_2
t_1>t_2+t_3. Then
(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3)-((t_2+t_3)+t_2+t_3)(1/(t_2+t_3) +1/t_2+1/t_3) =(t_1-t_2-t_3)(1/t_2+1/t_3-1/t_1) \geq 010>(t_1+t_2+t_3)(1/t_1+1/t_2+1/t_3) \geq 2(t_2+t_3)(1/(t_2+t_3)+1/t_2+1/t_3) <=> 4>(t_2+t_3)(1/t_2+1/t_3) \geq 4
x_1,x_2,...,x_nverify
k(n)>(x_1+x_2+...+x_n)(1/x_1+1/x_2+...+1/x_n)then any three of them are sides of a triangle.
(n+\sqrt{10}-3)^2. I think this one is much more interesting.
f(x_1,x_2,...,x_n)=(x_1+x_2+...+x_n)(1/x_1+1/x_2+...+1/x_n)where
x_1<=x_2<=...<=x_n.
k(n)=(n+\sqrt{10}-3)^2might be found elementary. From the solution of Sung-yoon Kim:
n^2+1>(\sqrt((x_1+x_2+x_3)(1/x_1+1/x_2+1/x_3))+(n-3)^2(x_1+x_2+x_3)(1/x_1+1/x_2+1/x_3)<(\sqrt(n^2+1) -(n-3))^2(n^2+1), use that
k(n)and instead of
x_3, use
x_n:
(x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)<(\sqrt(k(n)) -(n-3))^2x_1+x_2<=x_nthen:
(x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)>=10, so we must have
10>=(\sqrt(k(n)) -(n-3))^2,(x_1+x_2+x_n)(1/x_1+1/x_2+1/x_n)<(\sqrt(k(n)) -(n-3))^2, then:x_1+x_2>x_n, so for any
1<= i, j, k <= n:
x_i+x_j>x_k.
k(n)=min f(x_1,x_2,...,x_n),
x_1<=x_2<=...<=x_nand
x_1+x_2<=x_n, we may get:
k(n)=f(1,1,2\sqrt(2/5),...,2\sqrt(2/5),2), where
2\sqrt(2/5) appears (n-3)times.
Medges ,
M>=3so if the same inequality holds for any
n>=Mpositive reals (of course with something like
k(n,M)in the RHS) then any
Mfrom our
nreal numbers can be the side lengths of a polygon with
Medges. Solution is basically the same.
.
,
.
.
do not form the sides of a triangle. Then
.
for 
means that 
.
.
.
so it suffices to prove that 
,
.
implies that 
.
as desired.
,
such that 
and 
by AM-GM so 
as desired.
.
as desired.
,
,
where 
is nonnegative. By Cauchy-Schwarz, we have
Now
by AM-GM. To reach contradiction it is now sufficient to show 
which is true. We are done.
.
Thus, we want
if 
Since 
,
which is clearly a contradiction. Thus, for all 
,
and 
are side lengths of a triangle which implies that 
,
and 
are as well. Swapping the indices we get the conclusion for all 
,
,
and 
.
.
and do a second degree equation in 
.
,
so 
which is a contradiction from Cauchy.
.![\[t_1 + t_2 + t_3 = \frac25 t_1 + \frac35 t_1 + (t_2 + t_3) \geq \frac25 t_1 + \frac85(t_2 + t_3)\]](http://latex.artofproblemsolving.com/0/a/c/0acd6d3ef414f23102f5794365916ac04db3c1c9.png)
and ![\[\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} \geq \frac{1}{t_1} + \frac{4}{t_2 + t_3}\]](http://latex.artofproblemsolving.com/a/7/2/a7235fbb9869141c281b54ebbd1937a89bad445e.png)
by Cauchy-Schwarz, so we have ![\[(t_1 + t_2 + \dots + t_n)\left(\frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}\right)\]](http://latex.artofproblemsolving.com/7/6/6/76643350b294b5322309ee8358dca57f236afcb5.png)
![\[ \geq \left(\frac25 t_1 + \frac85(t_2 + t_3) + t_4 + \dots + t_n\right)\left(\frac{1}{t_1} + \frac{4}{t_2 + t_3} + \frac{1}{t_4} + \dots + \frac{1}{t_n}\right)\]](http://latex.artofproblemsolving.com/5/6/b/56be74ad5093a6bdb0b8c517fedd46f1846faee2.png)
![\[ \overset{\text{C-S}}{\geq} \left(\sqrt{\frac25} + \sqrt{\frac{32}{5}} + (n - 3)\right)^2 = \left(n + \sqrt{10} - 3\right)^2.\]](http://latex.artofproblemsolving.com/7/c/0/7c0c0364f68030796905b430e213165b3fd94381.png)
But 
is strictly increasing, so 
,
and 
?
'
.
,![\[f(x) = \sqrt x + 2\sqrt{2 - x},\]](http://latex.artofproblemsolving.com/c/3/7/c37b71145f5b1d1c15cfcfa44cbe556c1213ebbe.png)
which by C-S or similar methods is greatest when 
.
such that the inequality is satisfied and 
.
elements. Of course, just remove 
,
by AM-GM, so the inequality is preserved. Now we can just reduce to the 
.
are increasing for 
,
,
.
,
WLOG we can assume 
.
as parameters) we have:
We however know:
.
.
.
.
From this it directly follows that 
,
are the sides of a triangle for 
.
from which the desired will follow (because 
are the smallest). For simplicity let 
.
for some 
.
.
.
From here it follows that:
![\[\left(t_1 + t_2 + \dots + t_n\right)
\left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right) = n+\sum_{i, j \in \{1, 2, \dots, n\}, i\neq j} \frac{t_i}{t_j} +\frac{t_j}{t_i}\]](http://latex.artofproblemsolving.com/6/0/3/603a241e4653fd3df03efa30e822979ccf517f55.png)
Thus we get if for any 
we get a contradiction,
that 
,
,![\[\frac{1}{a}+\frac{1}{b}-\frac{a+b}{c(c+k)}\geq 0\]](http://latex.artofproblemsolving.com/d/6/3/d6347eedbee72e919170ae49d15d63dc6872409c.png)
When ![\[\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\geq 0\]](http://latex.artofproblemsolving.com/e/6/e/e6e5ffd6f245a148efe013b7f83b0e5e68252740.png)
![\[\frac{b+a}{ab}-\frac{1}{a+b}\]](http://latex.artofproblemsolving.com/4/9/7/497c7d2bc149121cfb147ceba6006cffeb0b1315.png)
The above is minimised when 
,![\[\frac{2a}{a^2}-\frac{1}{2a}\geq 0\]](http://latex.artofproblemsolving.com/3/d/9/3d9bcb384a9e7159c50d6748733c5ecf1e01c2f4.png)
Which is clearly true.
,
.
where the summation is taken over all ordered pairs in 
has no integer solutions for 
by AMGM. Also, 
by Cauchy so this reduces into 
which is obvious.
that are not the sides of a triangle, then 
For 
we must show that if 
WLOG, assume that 
then we have that 
Also, because our inequality is homogenous, assume that 
Letting 
it suffices to show that 
But by AM-GM, 
so it suffices to show that 
But, 
by 
Now, assume that for 
our proposition holds. Consider the sequence 
of positive real numbers. Then if there is at least one triple, say WLOG 
which are not the sides of a triangle then we have by our inductive hypothesis that
Therefore, our induction is complete. QED
But by our induction above, we have that 
a contradiction. Therefore, all 
Mainly for storage.
.
.
,
and 
.
.
.![$(s+u+v)(\frac{1}{s} + \frac{1}{u} + \frac{1}{v}) \ge (s+u+v)(\frac{4}{s+u} + \frac{1}{v}) = 5 + \frac{4v}{s+u} + \frac{s+u}{v} \ge 5 + 5 \cdot \sqrt[5]{\frac{v^3}{(s+u)^3}} \ge 10$](http://latex.artofproblemsolving.com/3/2/2/322de8dea08b6057f70ea16e1396449a2cc2af43.png)
when 
.
We rephrase the problem as such:
be a table with 
the number 
is written. A table is "good" if all of the numbers in it sum up to a number strictly less than 
.
for all 
.
written term by term:![\[
\begin{array}{c|ccccccc}
& \frac{1}{t_1}
& \frac{1}{t_2}
& \frac{1}{t_3}
& \frac{1}{t_4}
& \frac{1}{t_5}
& \cdots
& \frac{1}{t_n} \\
\hline
t_1 & & & & & & & \\
t_2 & & & & & & & \\
t_3 & & & & & & & \\
t_4 & & & & & & & \\
t_5 & & & & & & & \\
\vdots & & & & & & & \\
t_n & & & & & & & \\
\end{array}
\]](http://latex.artofproblemsolving.com/2/2/c/22c8c239fc3d7d4edd456b3ca64c7d274d2fa718.png)
Let us now start filling in the table with numbers - we immediately notice that the cells along the main diagonal all have the number 
inside of them, so we can immediately fill these ones up:![\[
\begin{array}{c|ccccccc}
& \frac{1}{t_1}
& \frac{1}{t_2}
& \frac{1}{t_3}
& \frac{1}{t_4}
& \frac{1}{t_5}
& \cdots
& \frac{1}{t_n} \\
\hline
t_1 &1 &a &b &c &d & &v \\
t_2 &a' &1 &g &h &i & &w \\
t_3 &b' &g' &1 &l &m & &x \\
t_4 &c' &h' &l' &1 &p & &y \\
t_5 &d' &i' &m' &p' &1 & &z \\
\vdots & & & & & &\ddots & \\
t_n &v' &w' &x' &y' &z' & &1 \\
\end{array}
\]](http://latex.artofproblemsolving.com/b/a/0/ba08e85c9ac36fc7a91ac7b4c59b0528aa25db6f.png)
Now we observe the following
etc.)
for all real 
.
with cell 
)
,
cells (two by two symmetric in regards to the main diagonal) with a sum of the numbers in them at least 
such that 
- assuming that such exists would invariably lead to a contradiction, thus completing the problem.
.
(note that if there exists a triple that does not form a triangle, then the triple 
also does not form one) and we have thus reduced the problem to proving that
Since the expression on the left is scalable, we can assume that 
,
and 
.
From here, we use the useful lemma 
(can be derived - pun intended - from the second derivative of the LHS function and proving that it is reached for 
) to get the inequality we need to prove to be simply
Since this function is increasing for 
(this can be verified by expressing the difference 
)
at 
,
cells with a sum of their numbers at least 
,
Then, we have that:
Thus, all that is left is to prove that 
Expanding, we have that 
Now, differentiate w.r.t. 
We get 
We claim that 
However, this is clearly true because 
while 
Thus, 
is increasing as 
This clearly implies that 
Hence, we are done.