AoPS Academy
be a trapezoid with parallel sides 
. Points 
and 
lie on the line segments 
and 
, respectively, so that 
. Suppose that there are points 
and 
on the line segment 
satisfying ![\[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]](http://latex.artofproblemsolving.com/b/1/a/b1ae208955104081bafe221832509d9dd20eeb83.png)
Prove that the points , , 
and 
are concyclic.
be a positive integer. A board with a format 
is divided in equal squares.Determine all integers ≥3 such that the board can be covered in 
(or 
) pieces so that there is exactly one empty square in each row and each column.
positive integers 
such that 
for any 
, 
such that 
?
denote the set of positive integers. Find all functions 
such that for positive integers 
and 
![\[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]](http://latex.artofproblemsolving.com/c/1/f/c1f8ffe04cfcc45b498f3931a4796d5c56dc04d0.png)
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
, the digits in the base-
representation of 
are all greater than .
be an equilateral triangle, and point 
on its circumcircle. Let 
and 
intersect at 
, 
and 
intersect at 
, and 
and 
intersect at 
. Prove that the area of 
is twice the area of 
.
, where 
and are relatively prime positive integers. Find 
.
for which 
is a perfect square.
be the set of real numbers. Let 
be a function such that ![\[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]](http://latex.artofproblemsolving.com/5/d/3/5d3476fd9195d085d0030d6036a9fde1cf6756a7.png)
for every 
. Assume that the inequality is strict for some 
.
for every 
or 
for every .
gives 
, which implies the result unless 
; henceforth suppose so.
produces 
or 
; but swapping 
analogously yields 
, and so 
.
gives
and chaining the 
inequality with the negation of the 
produces
where in the last step we only redenote 
for simplicity.
is strict for some 
: ![\[f(\tilde{w})f(\tilde{z})>-f(\tilde{w})f(-\tilde{z}).\]](http://latex.artofproblemsolving.com/b/c/2/bc221960553a1d7a4e262e8598929427a0903448.png)
But 
implies that the magnitudes of both sides are equal, and so both 
and 
are strictly positive. In particular, 
.
in yields 
, i.e., each 
has the same sign as 
. 
Co-authored with Ivan Chan (who proposed IMO 2023 P3 which imho a lot better than this), and Tristan Chaang (who was the first to supplya clean and neat solution; the original solution by Ivan and me made it looked like it's an A5).
, and then realized that leads to equality (i.e. no room on what to do). Then I decided to try the other direction and took me a while to notice the fact that became the problem statement!
and 
different signs? How about the case where 
for all 
? Those characterization seems elusive. )
. Here is the previous solution if anyone wants. 
denote the assertion.
.
, then we can just divide both sides by 
and switch 
to get that 
or 
for all 
.
.
.
in the above inequality, we get that 
. Adding these two up, we get 
which forces that equality holds in both the cases, i.e., 
for all .![\[
P(x,y) \implies f(x+y)f(x-y) \ge f(x)^2 - f(y)^2.
\]](http://latex.artofproblemsolving.com/9/6/3/9639b30c236f170e8905537382be3ade655774cb.png)
![\[
P(-y,x) \implies f(-y + x)f(-y-x) \ge f(-y)^2 - f(x)^2 = f(y)^2 - f(x)^2.
\]](http://latex.artofproblemsolving.com/9/5/b/95b5401f99fa8005a1c464ffbedf004430c2f9a1.png)
Adding these two up, we get that,![\[
f(x-y)(f(x+y) + f(-(x+y))) \ge 0
.\]](http://latex.artofproblemsolving.com/7/c/7/7c7d87681923af7c4bf10d4acfe24f0b2b62bd8c.png)
and 
to get,![\[
f(p)(f(q) + f(-q)) \ge 0
.\]](http://latex.artofproblemsolving.com/2/0/5/205ad9aa2bd6f94db02102fe438eae0ea9b27a99.png)
for some 
, then we are basically done. FTSOC assume that 
for all , i.e., 
is odd.![\[
P(x,y) \implies f(x+y)f(x-y) \ge f(x)^2 - f(y)^2
.\]](http://latex.artofproblemsolving.com/3/c/7/3c78c90440ce740103eb9fedc7f87f5861c8fc1b.png)
![\[
P(y,x) \implies f(y+x)f(y-x) \ge f(y)^2 - f(x)^2 \implies -f(x+y)f(x-y) \ge f(y)^2 - f(x)^2
.\]](http://latex.artofproblemsolving.com/5/e/2/5e2a3db0bc5133157a5cce426bef7f3997f51a73.png)
which forces the equality of both the equations, that is![\[
f(x+y)f(x-y) = f(x)^2 - f(y)^2
\]](http://latex.artofproblemsolving.com/a/2/4/a24437641d82c49de1a8098c8dfbb70c4dc1034f.png)
for all , 
. But this is a contradiction because for 
the equality fails and we are done.
means?
denote the given assertion. First, 
gives 
for all . Thus, we are done if 
. Now, assume . Then, 
and 
give![\[f(-x)^2 \geq f(x)^2 \geq f(-x)^2 \implies f(x)^2 = f(-x)^2 \enspace \forall x \in \mathbb{R}.\]](http://latex.artofproblemsolving.com/7/6/e/76e6d667dcc1acf83c08b10859ebc82237f5783c.png)
We have
for all 
, and denote this by 
. Suppose 
for some . Then 
and 
yield
Again, replacing by 
gives 
. It follows that 
for all real , assuming that 
. Then, for any ,
which forces to be the equality, contradicting the hypothesis.
for all . It follows that
That is, 
for all real . Now, the conclusion follows.
be a pair that makes the inequality strict. Plugging in the pairs and 
and summing gives us![\[ f(a+b) [ f(a-b) + f(b-a) ] > 0 \implies f(a-b)+f(b-a) \neq 0.\]](http://latex.artofproblemsolving.com/3/9/4/394e531034cb3855514dd0d5c62b4ca8c74bb98e.png)
Let be any constant; plugging in the pairs 
and 
and summing gives us![\[ f(a+b+2k) [ f(a-b) + f(b-a) ] \geq 0 \implies \text{sgn} (f(a+b+2k)) = \text{sgn} (f(a-b)+f(b-a))\]](http://latex.artofproblemsolving.com/9/a/d/9ad122276ded880dca8e6b662cd7b68d3fd4173b.png)
Since varies, 
can be any real number, so we're done.
achieving both signs implies that the inequality is an equality.
and 
are of different signs (if they are not both zero)
be such that and 
have different signs. Then choose and such that 
and 
. Then gives that 
so 
gives that 
must be positive, as desired.
then 
would contradict our assumption.
and 
. and shows that the inequality must be an equality.
denote the assertion.![\[-f(x+y)f(y-x) \le f(x)^2 - f(y)^2 \le f(x+y)f(x-y)\]](http://latex.artofproblemsolving.com/5/3/5/53513c602b0673c5b6b45b4fd1b75bcb010b9772.png)
![\[-f(x+y)f(y-x) \le f(x+y)f(x-y)\]](http://latex.artofproblemsolving.com/f/1/b/f1b1bee4f64309bcb67a12449159271fd3e5fc60.png)
such that 
, then we have that by 
, which gives that 
which implies that 
has the same sign as 
which implies or .
for all 
, then we get that 
so the inequality is never strict.
denote the assertion
and 
give
and
respectively. We add them to get
and 
give
and
respectively. We add them to get
so
or
for all as desired, and we are done. 
be the assertion of 
.
inequality yields :![$f(x_0+y_0)[f(x_0-y_0)+f(y_0-x_0)]>0$](http://latex.artofproblemsolving.com/2/9/1/291a5afe00d7b5176280cabde6663fe51f8f8a6a.png)
...
then 
for all 
we get 
.
thus 
for all .
thus 
for all 
. then
. 
we have
or 
..
, contradiction. Therefore, .
such that 
, then 
or 
for all 
inequality yields :![$f(b)[f(a)+f(-a)]\ge 0$](http://latex.artofproblemsolving.com/8/c/3/8c32864872b63a79abc688d03ca130030c397b1d.png)
then 
. If 
then 
. . We claim that 
.
.
.
such that . Using Claim , we are finished. Done. 
is a solution then 
is a solution. As such, WLOG scale such that 
.. By 
we get that 
which finishes if 
, so assume . we get that ![\[ 0 \ge f(x)^2 - f(-x)^2 \]](http://latex.artofproblemsolving.com/9/1/7/91759e2e4f154e0e7d2ef75e22d452f5d673c8b4.png)
Comparing with , we get that . is either even or odd.
. Then set 
. Now 
have LHS with opposite signs so 
. By comparing 
we get 
as well.
and thus 
, contradiction. is odd. Note that by 
we have that 
. Since 
it follows that equality must hold for all , contradiction. is even. Take 
such that 
. has a negative sign. Then take 
.![\[ f(a) \ge f(x)^2 - f(y)^2 \]](http://latex.artofproblemsolving.com/3/4/e/34ed95b4c10ed4d72acf2fe9ce83fbc2ad9ea2b2.png)
By , we also get ![\[ f(a) \le f(x)^2 - f(y)^2 \]](http://latex.artofproblemsolving.com/c/c/9/cc9eaa95a6d9c5c73a8f9b10a983c80b9e3fb9f0.png)
This gives a contradiction.
and different signs? How about the case where for all ? Those characterization seems elusive. )
denote the given assertion. Suppose that the result was not true and choose reals 
such that 
and 
. Additionally, choose 
such that 
.
. However, choosing 
implies that 
, so 
, but 
implies that 
, so 
, absurd. Thus, must equal 
. 
. But 
gives 
, so this means , so either 
or .
. Notice that and 
gives that 
. Now, taking the negative of both sides and flipping the inequality gives that 
, absurd. 
.
.
, then 
and 
, so 
. However, if 
, then 
, contradiction to our claim that . Therefore, 
, choosing 
such that 
and 
have opposite signs gives that 
, contradiction. always or always.
as the assertion to the inequality. With the usual swapping trick, consider and . This will yield the following two inequalities.
Add them up and write 
for some arbitrary real .![\[f(2x+a) \left(f(-a) + f(a) \right) \ge 0.\]](http://latex.artofproblemsolving.com/4/2/6/426091d8b1eaf59f04d1073ef661b283e8711232.png)
If is not an odd function, then we can find some 
such that 
. Then dividing the above inequality by 
with an appropriate change in sign of inequality, we would be done. Henceforth assume is an odd function. is odd, 
. From the above aligned equations, we can now say that![\[f(x+y)f(x-y) = f(x)^2 - f(y)^2 \]](http://latex.artofproblemsolving.com/7/0/5/70547d122bfb3e4df2a25f411a5823330ad4e12c.png)
for all 
which is the desired contradiction since there exists some 
such that there is strict inequality in . This completes the solution. 
denote the given assertion.
If is not 0, then the problem statement is directly implied, Hence, suppose 
, replacing by 
. Thus, 
. Now, Note that the problem condition can be translated to 
for every 
for all real 
(as ) exists such that 
. and such that and 
is not identically 0; since satisfies the FE iff 
satisfies the FE, WLOG let 
, and denote the FE as .
, so either in which we're done, or .
. But as well so 
.
, and if we choose such that 
and 
, then 
for all 
, which by 
implies 
or 
. Let the set of all 
that satisfy the latter be ; note that if 
we're done.
satisfies 
, then 
. But 
already so 
. But substituting 
, we similarly get 
. If we choose to further specify that 
, adding with the previous equality, gives 
as 
. Thus, if 
, then 
for all .
, that 
for all 
; yet the problem statement says there should exist for which this is not true. Thus, and we conclude for all . 
gives us 
. This gives , if not, then for all , the sign of doesn't change.
gives 
, changing with 
gives us 
, which implies 
and . as gives 
. and gives 
which means 
.
. Choose and as , 
, where c is any real number.
Putting 
as 
gives 
which implies 
.
, 
.
, which gives us contradiction. 
, 
. Thus, if , we may assume without loss of generality that , from where 
for all real numbers , so the desired result is true., then ![\[f(x)^2-f(-x)^2 \leq f(0)f(2x) = 0.\]](http://latex.artofproblemsolving.com/9/1/b/91b3bd4ae599cd78a9c78b045e7f312b42cdffe8.png)
In particular, swapping and , it follows that . Now, summing the equations for 
and 
,![\[f(x+y)(f(x-y) + f(y-x)) \geq f(x)^2-f(y)^2+f(y)^2-f(x)^2 = 0.\]](http://latex.artofproblemsolving.com/a/8/e/a8e273c40df4d8815a19ab1893844481a17de6bd.png)
In particular, there is 
such that this inequality is strict; i.e. for this pair , it follows that 
where 
. However, for all pairs 
, it follows that![\[f(2x+c)(f(c)+f(c))) \geq 0\]](http://latex.artofproblemsolving.com/e/8/7/e87e505b7fc859f4e238edf7af8f1fa700964bcb.png)
by the same argument, hence if 
, 
for all and vice versa.
with 
, equality must always hold.. to get 
If , we're done as equality holds and thus must hold in the original equation.
, choosing such that comes out to the value required and 
, we have 
. But doing the same such that 
, 
.Thus 
, which would mean (yet again) that equality holds. Thus we're done,
or the problem is true.
we get
so if 
then all of 
have the same sign, and we are done.
we get 
so 
Also, 
so 
as desired.
and adding these together, we get
and we will call this assertion 
for some 
then we are done.
gives
Since was arbitrary, we are done. 
satisfies 
Then, 
would be at an equality case. However, since is the sum of two inequalities, one of which is 
which is not an equality case, we are done. 
. and gives us 
.
, there exists such that 
and 
.
for any 
.
., 
is also strict for 
. such that 
.
. This also implies 
.
in gives 
.
can be done in a similar way.
we get that 
and so we are done if . Suppose . Putting we get 
and so, since this holds for all , 
.
with 
(in particular, note that 
). Putting 
, 
we get 
and putting 
, 
we get 
Thus, 
for all and we're done.
for all real . Putting 
we get 
thus 
for all 
. But this is a contradiction with the assumption that the inequality is strict for some 
.
Plug in 
and add to get 
Since there exist 
such that the inequality is strict, we have 
or 
The two cases are the same, so we assume it is positive. Plug in 
and we get 
, we get 
. If , this implies and have the same sign, which finishes the problem.. Taking 
, we get 
, from which 
. Replacing with , we get 
, and from the two inequalities we get 
.
. Replacing and in the statement, we get 
, and by summing with 
, we have ![\[f(x+y)(f(x-y) + f(y-x)) \geq.\]](http://latex.artofproblemsolving.com/1/a/4/1a47e2fcca214a867b6aa53708b64609a031c71b.png)
We know there exist for which the inequality in the statement is strict, which means there exist for which 
, which implies 
, otherwise their sum is null. Let 
. Then 
, for any 
with 
. Let 
, then 
, so 
and 
have the same sign, which solves the problem, as 
takes any real value.
denote assertion
we are done, assume 
From this we can see that 
has same sign as 
or is equal to 0
then 
is not nonnegative or nonpositive. is obviously not universally zero..
, then , if the expression on the left can have either sign, so .
, then 
, so . is odd. with 
, which is possible by our assumption. Then 
, so 
, so 
is positive. Now take any 
, if 
, set 
with 
, 
, thus 
. Then 
, so 
is negative. Combining this with the above claim gives 
for 
. If 
, set 
, then 
, then , so is positive, combining this with the above claim gives 
for all values of 
, thus it is always true.
gives 
, forcing 
, implying that the inequality is never strict, giving the desired contradiction.
be the assertion 
.
.
(by letting 
), we have 
. was identically zero, i.e., 
for all , then 
, which is 
. This holds with equality for all . However, the problem states that the inequality is strict for some 
. This means cannot be identically zero... then becomes 
, which is . This provides no information on its own.
gives 
, also equality if .
gives 
. and satisfies 
(i.e., equality everywhere), then typical solutions (often requiring continuity, though this problem doesn't state it) are 
, 
, or 
. For these functions, the equality holds for all . and the given inequality were to hold with equality for all , this would contradict the given condition that the inequality is strict for some .: if , then the original inequality for 
would be 
, i.e., 
. This can't be the strict inequality. So 
. Similarly for 
, 
. If this were the strict case 
, it doesn't force .
(for 
) gives the strict inequality, 
. This would directly imply . The problem states "for some ". If this pair can be 
, then . If it cannot be (because 
for all ), then (as is not identically zero). This implies 
must hold with equality for all to satisfy when , which contradicts strict inequality..: for all :, then 
for all , then 
for all 
denote the given inequality.
gives 
If 
then we are automatically done. Henceforth assume 
for all 
to get 
and use to get 
Comparing the two gives 
as claimed. such that 
and along with the given condition to get
If 
then that gives 
a contradiction. Hence we must have and 
so using 
suffices.
and 
give
Adding the two gives 
If 
then we get for all If 
then we get for all Therefore, we are done.
and 
Prove that
If 
is strict, then 
gives 
so 
gives 
If 
then 
then 
part from the inequality which can't be done if 
because we are given that the inequality is strict for some 
.
.![\[ P(0,y) \implies f(y)f(-y) \ge -f(y)^2 \implies f(y)^2 \ge -f(y)f(-y) \implies f(y)(f(y) + f(-y)) \ge 0. \]](http://latex.artofproblemsolving.com/9/7/3/9735fb7cb4b210681510d2a41bf5c28d32ff464a.png)
in the above equation, we get that,![\[ f(-y)(f(-y)+f(y)) \ge 0. \]](http://latex.artofproblemsolving.com/f/a/3/fa3b677534966c8c23998bd8846b96cce7a0cb5d.png)
Now we can multiply the two inequalities to further get,![\[ f(y)f(-y)(f(y) + f(-y)^2 \ge 0\]](http://latex.artofproblemsolving.com/3/b/c/3bcb2497031b717c67c8286a5329591bdf2e59a8.png)
which further gives that 
for all 
and 
have the same sign.![\[ P(x,-x) \implies 0 \ge f(x)^2 - f(-x)^2 \implies f(-x)^2 \ge f(x)^2. \]](http://latex.artofproblemsolving.com/2/5/0/250571f6e8b61dd22619257f4fc0c823f4764e33.png)
![\[ f(-x)^2 + f(x)^2 \ge f(x)^2 + f(-x)^2 \]](http://latex.artofproblemsolving.com/c/b/b/cbbceab9e9a511c2513d255b29b17d095853c70c.png)
which forces that equality holds in both the cases, i.e., 
.![\[ P(x,y) \implies f(x+y)f(x-y)\ge f(x)^2-f(y)^2 \]](http://latex.artofproblemsolving.com/2/9/c/29c8c9bdb85db055c5a165352a6de417b970da1f.png)
![\[ P(y,x) \implies f(x+y)f(y-x) \ge f(y)^2 - f(x)^2 \implies f(x+y)f(x-y) \ge f(y)^2 - f(x)^2. \]](http://latex.artofproblemsolving.com/2/3/5/2359b53f6353941f06e51ccebcae7617d721ce79.png)
Adding these two up, we get 
.
in 
which clearly forces 
. Will be posting soon.
,
into the functional equation. Combining the strict 
Note that 
where 
.
and 
now shows 
,
.
- let this be 
.
using 
for any reals 
The existence of 
.
and varying 
implies that 
we get 
.
Notice that if we have equality in 
we must have 
so 
or 
is 
we get (keep in mind that the inequality is strict) 
.
.
we will show that 
.
and 
.
we have equality and since 
we get 
.
hence 
.
we will show that 
.
since 
so 
.
![\[P(x_0, y_0)+P(y_0, x_0)\]](http://latex.artofproblemsolving.com/1/9/3/19359150768f129ddff3dad1e1ad03a7c6a16637.png)
![\[f(x_0+y_0)(f(x_0-y_0)+f(y_0-x_0))>0\]](http://latex.artofproblemsolving.com/d/c/c/dcc1e263b816f10fd46ecc30121d1bb0fd8da3d3.png)
Let 
.![\[P(x, x+s)+P(x+s, x)\]](http://latex.artofproblemsolving.com/7/4/6/746229a31097c786b25ebc67792c8c2584fa32ca.png)
![\[f(2x+s)(f(s)+f(-s))\geq 0\]](http://latex.artofproblemsolving.com/4/f/5/4f57d5fa6001dbb783ea3363332ceaa92d80e42d.png)
Thus for all 
.
gives us![$$f(a + b)[f(a - b) + f(b - a)] \ge 0 \;\;\;\; \dots (\clubsuit)$$](http://latex.artofproblemsolving.com/7/0/b/70bb87c5880f2c93ac4169bed2d791d5403c1741.png)
Let the assertion in 
be 
gives us![$$f(x_0 + y_0)[f(x_0 - y_0) + f(y_0 - x_0)] > 0$$](http://latex.artofproblemsolving.com/0/e/b/0eb3c42ea151ea2872420f2c29566168764bda71.png)
with a strict inequality sign since one of 
and 
are both nonzero and of the same sign. Now, 
gives![$$f(x_0 + y_0 + 2k)[f(x_0 - y_0) + f(y_0 - x_0)] \ge 0$$](http://latex.artofproblemsolving.com/3/1/a/31a81f5abefcd1bd560a152b0fc43aaca5a95e79.png)
Since 
(or the latter quantity is zero, which does not matter). Since 
.
.![$f(x+y)[f(x-y)+f(y-x)] \geq 0$](http://latex.artofproblemsolving.com/e/0/d/e0d4e081d459b3bba12a7f6bd0e9effbfe9ff5e0.png)
.
,![$f(a)[f(b)+f(-b)] \geq 0$](http://latex.artofproblemsolving.com/3/b/e/3beb08afb4e3d825c40f816a85f185708f5c4913.png)
,
.
such that 
.