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Inspired by Bet667
sqing   4
N 10 minutes ago by ytChen
Source: Own
Let $ a,b $ be a real numbers such that $a^3+kab+b^3\ge a^4+b^4.$Prove that
$$1-\sqrt{k+1} \leq  a+b\leq 1+\sqrt{k+1} $$Where $ k\geq 0. $
4 replies
sqing
Thursday at 1:03 PM
ytChen
10 minutes ago
Inspired by Kosovo 2010
sqing   2
N 12 minutes ago by sqing
Source: Own
Let $ a,b>0  , a+b\leq k $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq\left(1+\frac{4}{k(k+2)}\right)^2$$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq\left(1+\frac{2}{k+2}\right)^2$$Let $ a,b>0  , a+b\leq 2 $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq \frac{9}{4} $$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq \frac{9}{4} $$
2 replies
sqing
Yesterday at 3:56 AM
sqing
12 minutes ago
Triangle on a tetrahedron
vanstraelen   2
N Yesterday at 7:51 PM by ReticulatedPython

Given a regular tetrahedron $(A,BCD)$ with edges $l$.
Construct at the apex $A$ three perpendiculars to the three lateral faces.
Take a point on each perpendicular at a distance $l$ from the apex such that these three points lie above the apex.
Calculate the lenghts of the sides of the triangle.
2 replies
vanstraelen
Yesterday at 2:43 PM
ReticulatedPython
Yesterday at 7:51 PM
shadow of a cylinder, shadow of a cone
vanstraelen   2
N Yesterday at 6:33 PM by vanstraelen

a) Given is a right cylinder of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the botom base?

b) Given is a right cone of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the base?
2 replies
vanstraelen
Yesterday at 3:08 PM
vanstraelen
Yesterday at 6:33 PM
2023 Official Mock NAIME #15 f(f(f(x))) = f(f(x))
parmenides51   3
N Yesterday at 5:13 PM by jasperE3
How many non-bijective functions $f$ exist that satisfy $f(f(f(x))) = f(f(x))$ for all real $x$ and the domain of f is strictly within the set of $\{1,2,3,5,6,7,9\}$, the range being $\{1,2,4,6,7,8,9\}$?

Even though this is an AIME problem, a proof is mandatory for full credit. Constants must be ignored as we dont want an infinite number of solutions.
3 replies
parmenides51
Dec 4, 2023
jasperE3
Yesterday at 5:13 PM
Geometry
AlexCenteno2007   3
N Yesterday at 4:18 PM by AlexCenteno2007
Let ABC be an acute triangle and let D, E and F be the feet of the altitudes from A, B and C respectively. The straight line EF and the circumcircle of ABC intersect at P such that F is between E and P, the straight lines BP and DF intersect at Q. Show that if ED = EP then CQ and DP are parallel.
3 replies
AlexCenteno2007
Apr 28, 2025
AlexCenteno2007
Yesterday at 4:18 PM
Cube Sphere
vanstraelen   4
N Yesterday at 2:37 PM by pieMax2713

Given the cube $\left(\begin{array}{ll} EFGH \\ ABCD \end{array}\right)$ with edge $6$ cm.
Find the volume of the sphere passing through $A,B,C,D$ and tangent to the plane $(EFGH)$.
4 replies
vanstraelen
Yesterday at 1:10 PM
pieMax2713
Yesterday at 2:37 PM
Square number
linkxink0603   3
N Yesterday at 2:36 PM by Zok_G8D
Find m is positive interger such that m^4+3^m is square number
3 replies
linkxink0603
Yesterday at 11:20 AM
Zok_G8D
Yesterday at 2:36 PM
Combinatorics
AlexCenteno2007   0
Yesterday at 2:05 PM
Adrian and Bertrand take turns as follows: Adrian starts with a pile of ($n\geq 3$) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
0 replies
AlexCenteno2007
Yesterday at 2:05 PM
0 replies
How many pairs
Ecrin_eren   6
N Yesterday at 12:57 PM by Ecrin_eren


Let n be a natural number and p be a prime number. How many different pairs (n, p) satisfy the equation:

p + 2^p + 3 = n^2 ?



6 replies
Ecrin_eren
May 2, 2025
Ecrin_eren
Yesterday at 12:57 PM
parallelogram in a tetrahedron
vanstraelen   1
N Yesterday at 12:19 PM by vanstraelen
Given a tetrahedron $ABCD$ and a plane $\mu$, parallel with the edges $AC$ and $BD$.
$AB \cap \mu=P$.
a) Prove: the intersection of the tetrahedron with the plane is a parallelogram.
b) If $\left|AC\right|=14,\left|BD\right|=7$ and $\frac{\left|PA\right|}{\left|PB\right|}=\frac{3}{4}$,
calculates the lenghts of the sides of this parallelogram.
1 reply
vanstraelen
May 5, 2025
vanstraelen
Yesterday at 12:19 PM
Find max
tranlenhanhbnd   0
Yesterday at 11:50 AM
Let $x,y,z>0$ and $x^2+y^2+z^2=1$. Find max
$D=\dfrac{x}{\sqrt{2 y z+1}}+\dfrac{y}{\sqrt{2 z x+1}}+\dfrac{z}{\sqrt{2 x y+1}}$.
0 replies
tranlenhanhbnd
Yesterday at 11:50 AM
0 replies
easy functional
B1t   13
N Apr 30, 2025 by AshAuktober
Source: Mongolian TST 2025 P1.
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
13 replies
B1t
Apr 26, 2025
AshAuktober
Apr 30, 2025
easy functional
G H J
G H BBookmark kLocked kLocked NReply
Source: Mongolian TST 2025 P1.
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B1t
24 posts
#1 • 1 Y
Y by farhad.fritl
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
This post has been edited 2 times. Last edited by B1t, Apr 26, 2025, 7:01 AM
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NicoN9
148 posts
#2
Y by
$f(x)f(x)$ means $f(x)^2$? or is it a typo?
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Ilikeminecraft
624 posts
#3
Y by
$x = y = z = 0$ implies $f(0) = 0$
$x = -y$ implies $f(z) = f(z) - xf(x) + f(x)^2,$ so $f\in\{0, x\}.$
assume $f(a ) = 0, f(b ) = b.$
take $y = 0,$ and we get $f(xf(x) + z) = f(z) + f(x)^2.$
if we take $x = b, y = a - b, z = a,$ we get $0 = ab.$ this implies $f \equiv 0, x.$
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B1t
24 posts
#5
Y by
NicoN9 wrote:
$f(x)f(x)$ means $f(x)^2$? or is it a typo?

I wrote it wrong. im sorry
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Ilikeminecraft
624 posts
#6
Y by
suppose $f$ is not identically $0$
suppose $x_0$ satisfies $f(x_0)\neq 0,$ and if we pick $x = x_0$ and range the values of $y,$ we obtain that $f$ is surjective. ignore the definition of $x_0$
take $y = 0$ and we get $f(xf(x) + z) = f(xf(x)) + f(z)$
take $x = - y$ yields $f(xf(x))=xf(x).$
thus, $f(f(xf(x)) + z) = f(z) + f(f(xf(x)))$ which implies that the function is cauchy
plugging back into original equation, one gets $f(xf(y)) = f(x)y$
pick $x = 1$ to get $f(f(y)) = y,$ which is well-known to imply that $|f(y)| = |y|.$

assume $A$ is the set of $x$ such that $f(x) = x$ and $B$ is the set of $x$ such that $f(x) = -x.$
if $x, y \in A,$ then $f(xy) = xy,$ so $xy \in A$
if $x\in A, y\in B,$ then $f(xy) = -xy,$ so $xy\in B$
if $x \in B, y \in A,$ then $f(xy) = -xy,$ so $xy \in B$
if $x \in B, y\in B,$ then $f(xy) = xy,$ so $xy\in A$

this implies $f$ is multiplicative
hence, since $f$ is both multiplicative and additive, $f$ is identity
This post has been edited 1 time. Last edited by Ilikeminecraft, Apr 26, 2025, 3:22 PM
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Haris1
77 posts
#7
Y by
I wont write the solution , i will just write the steps.
$1.$ Prove that the function is additive
$2.$ Prove that its either constant or bijective
$3.$ Prove that its multiplicative and finish
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cazanova19921
552 posts
#8 • 1 Y
Y by farhad.fritl
B1t wrote:
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
P(x,y,z):\, f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
$P(0,0,0)$ $\implies$ $f(0)=0$.
$P(x,-x,0)$ $\implies$ $f(xf(x))=xf(x)$
So $P(x,y-x,z): \, f(xf(y)+z)=f(z)+yf(x)$ (new $P$)
- If $f(t)=0$ for some $t \neq 0$, then $P(x,t,0)$ $\implies$ $\boxed{f=0}$ which is a solution.
- Suppose $f(t)=0$ $\iff$ $t=0$.
$P(1,x,0)$ $\implies$ $f(f(x))=xf(1)$. hence $f$ is bijective, replace $x=1$ in this equation, we get $f(f(1))=f(1)$ so $f(1)=1$.
Therefore $f(f(x))=x$ for all $x$.
$P(x, f(y), 0)$ $\implies$ $f(xy)=f(x)f(y)$ for all $x, y$
$P(x, 1, y)$ $\implies$ $f(x+y)=f(x)+f(y)$ for all $x, y$
So $f$ is additive and multiplicative, hence $\boxed{f=\mathrm{Id}}$ which is also a valid solution.
This post has been edited 1 time. Last edited by cazanova19921, Apr 26, 2025, 1:55 PM
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MathLuis
1524 posts
#9
Y by
Denote $P(x,y,z)$ the assertion of the given F.E.
Notice from $P(0,y,z)$ we can inmediately get $f(0)=0$ and also $P(x,-x,z)$ gives that $f(xf(x))=xf(x)$ but also $P(x,y,0)$ now gives that $f(xf(x+y))=(x+y)f(x)$ so in fact shifting gives $Q(x,y)$ which is $f(xf(y))=yf(x)$ now notice if ther existed some $c \ne 0$ for which $f(c)=0$ then $Q(x,c)$ gives that $f(x)=0$ for all reals $x$ so its either that or $f$ is injective at zero, but now basically note that our F.E. may now by re-written as $R(x,y,z)$ to be $f(xf(y)+z)=f(z)+yf(x)$ for all reals $x,y,z$ but also take $y \ne 0$ and shift $x$ to get that $f$ is additive, but also from $Q(x, f(x))$ we get that $f(f(1)x^2)=f(x)^2$ and therefore shifting $x \to f(x)$ gives $f(f(1)f(x)^2)=f(1)^2 \cdot x^2$ so for example $f$ is surjective on all non-negative reals also from $Q(x,y)$ we have $f$ injective trivially when setting $x \ne 0$ and this take $f(d)=1$ and $Q(x,d)$ to get that $d=1$ and thus $f(1)=1$ so $Q(1,x)$ gives $f$ is an involution so by $Q(x,f(y))$ we get $f$ multiplicative so addivitive+multiplicative means $f$ is the identity function or constant (later case gives $f$ is zero everywhere so doesn't count), so $f(x)=0,x$ for all reals $x$ are the only two solutions that work thus we are done :cool:.
This post has been edited 1 time. Last edited by MathLuis, Apr 27, 2025, 3:32 PM
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jasperE3
11317 posts
#10
Y by
B1t wrote:
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]

Let $P(x,y)$ be the assertion $f(xf(x+y)+z)=f(z)+yf(x)+f(xf(x))$.
Note that $\boxed{f(x)=0}$ is a solution, else there is some $j$ with $f(j)\ne0$.
$P(0,0,0)\Rightarrow f(0)=0\Rightarrow j\ne0$
$P(j,-j,0)\Rightarrow f(jf(j))=jf(j)$
$P(j,0,x)\Rightarrow f(x+jf(j))=f(x)+jf(j)$
$P(x,y+jf(j),z)\Rightarrow f(xf(x+y)+jf(j)x+z)=f(z)+yf(x)+f(xf(x))+jf(j)f(x)$
Comparing this last line with $P(x,y)$ we get:
$$f(xf(x+y)+jf(j)x+z)=f(xf(x+y)+z)+jf(j)f(x)$$and setting $z=-xf(x+y)$ and $x=1$ this is $jf(j)(f(1)-1)=0$, so $f(1)=1$.
$P(1,x-1,y)\Rightarrow f(f(x)+y)=x+f(y)$
Call this assertion $Q(x,y)$.
$Q(x,0)\Rightarrow f(f(x))=x$
$Q(f(x),y)\Rightarrow f(x+y)=f(x)+f(y)$
$P(x,0,y)\Rightarrow f(xf(x)+y)=f(xf(x))+f(y)$
$P(x,-x,0)\Rightarrow f(xf(x))=xf(x)$
Now $P(x,y)$ becomes:
\begin{align*}
xf(x)+f(z)+f(xf(y))&=f(xf(x))+f(z)+f(xf(y))\\
&=f(xf(x)+xf(y)+z)\\
&=f(xf(x+y)+z)\\
&=f(z)+yf(x)+f(xf(x))\\
&=f(z)+yf(x)+xf(x)
\end{align*}and so $f(xf(y))=yf(x)$. Taking $y\mapsto f(y)$ we have $f(xy)=f(x)f(y)$, well-known that the only additive and multiplicative functions are $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ which both work.
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GreekIdiot
221 posts
#11
Y by
why is mongolian tst so easy?
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B1t
24 posts
#12
Y by
GreekIdiot wrote:
why is mongolian tst so easy?

true
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MuradSafarli
109 posts
#13 • 2 Y
Y by B1t, Nuran2010
Interesting problem!

Let \( P(x, -x, y) \) denote the assertion of the functional equation:
\[
xf(x) = f(xf(x)) \tag{1}
\]
Now, consider the following:

- \( P(1, -1, x) \) gives:
\[
f(1) = f(f(1)).
\]- \( P(1, x-1, 0) \) gives:
\[
f(f(x)) = x \cdot f(1). \tag{2}
\]
Now, we consider two cases:

---

**Case 1:** \( f(1) = 0 \)

From equation (2), we have:
\[
f(f(x)) = 0 \quad \text{for all } x.
\]
Applying \( f \) to both sides of equation (1):
\[
xf(x) = f(xf(x)) = f(f(xf(x))) = 0,
\]thus implying:
\[
f(x) = 0 \quad \text{for all } x.
\]
---

**Case 2:** \( f(1) \neq 0 \)

From equation (2), we can conclude that \( f \) is bijective.

Suppose there exists some \( k \) such that \( f(k) = 1 \).
Applying \( P(k, -k, x) \) gives:
\[
k = 1,
\]thus \( k = 1 \).

Moreover, from (2), we obtain:
\[
f(f(x)) = x.
\]
Now, consider \( P(1, f(x) - 1, y) \). We get:
\[
f(x) + f(y) = f(x+y),
\]meaning \( f \) is **additive**.

Since \( f \) is additive and satisfies \( f(f(x)) = x \), we deduce that \( f(x) = cx \) for some constant \( c \). Substituting back into the original functional equation shows that \( c = 1 \), and thus:
\[
f(x) = x.
\]
---

**Final answer:**
1. \( f(x) = 0 \) for all real numbers \( x \), or
2. \( f(x) = x \) for all real numbers \( x \).
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complex2math
7 posts
#14
Y by
Denote by $P(x, y, z)$ the assertion of the given functional equation. Then $P(0, 0, 0)$ and $P(x, -x, 0)$ gives the following

Claim 1. $f(0) = 0$ and $f(xf(x)) = xf(x)$.

Now we can rewrite the given functional equation as
\[
f(xf(x + y) + z) = f(z) + f(x)(x + y) \qquad (\heartsuit)
\]and let $P(x, y, z)$ re-denote the assertion of equation $(\heartsuit)$ and set $f(1) = a$.

Claim 2. If $a = 0$, then $f(x) \equiv 0$.

Proof. $P(x, 1 - x, z)$ implies $f(z) = f(ax + z) = f(z) + f(x)$, so $f(x) = 0$ for all $x \in \mathbb{R}$.

In what follows, we always assume $a \ne 0$.

Claim 3. $f(x)$ is bijective when $a \ne 0$.

Proof. From $P(1, y, 0)$ we get $f(f(1 + y)) = a(y + 1) \implies f(f(y)) = ay$. This is a bijection whenever $a \ne 0$.

Claim 4. $a = f(1) = 1$ and $f(x)$ is additive, i.e. $f(x + y) = f(x) + f(y)$ holds.

Proof. Since $xf(x) = f(xf(x))$, substituting $x = 1$ we obtain $f(1) = f(f(1)) \implies f(1) = 1$ as $f$ is injective. Then $P(x, 1 - x, z)$ gives $f(x + z) = f(z) + f(x)$.

Claim 5. $f(x)$ is multiplicative, i.e. $f(xy) = f(x)f(y)$ holds.

Proof. We have $f(f(y)) = ay = y$ so $P(x, y, 0)$ gives $f(xf(x + y)) = f(x)(x + y) = f(x)\cdot f(f(x + y))$. Then note that
\[
S_x := \{f(x + y): y \in \mathbb{R}\} = \mathbb{R}
\]for fixed $x \in \mathbb{R}$ since $f$ is surjective.

It's well-known that if $f(x)$ is both additive and multiplicative, then either $f(x) \equiv 0$ or $f(x) = x$.
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AshAuktober
1005 posts
#15
Y by
The claims I made in order to solve:
1) $f(0) = 0$.
2) $f(xf(x)) = xf(x)$.
3) $f(xf(y)) = yf(x)$.
4) Either $f \equiv 0$ or $f(x) = 0 \implies x = 0$.
(Now onwards taking second case...)
5) $f$ is an involution
6) $f$ is additive
7) $f$ is multiplicative
And done!
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