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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
IMO ShortList 1999, geometry problem 2
orl   13
N 18 minutes ago by ezpotd
Source: IMO ShortList 1999, geometry problem 2
A circle is called a separator for a set of five points in a plane if it passes through three of these points, it contains a fourth point inside and the fifth point is outside the circle. Prove that every set of five points such that no three are collinear and no four are concyclic has exactly four separators.
13 replies
+1 w
orl
Nov 13, 2004
ezpotd
18 minutes ago
Brilliant Problem
M11100111001Y1R   5
N 38 minutes ago by Davdav1232
Source: Iran TST 2025 Test 3 Problem 3
Find all sequences \( (a_n) \) of natural numbers such that for every pair of natural numbers \( r \) and \( s \), the following inequality holds:
\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]
5 replies
+1 w
M11100111001Y1R
May 27, 2025
Davdav1232
38 minutes ago
4 var inequality
SunnyEvan   0
42 minutes ago
Source: Own
Let $ x,y,z,t \in R^+ ,$ such that : $ (x+y+z+t)^2 = x+y+z+t + (x+z)(y+t) $ and $ x \geq y \geq z \geq t .$
Try to prove or disprove : $$ \frac{2 \sqrt{x+y+z+t +(x+t)(y+z)}}{x^2+y^2+z^2+t^2 +3xz+3yt+xt+yz} \geq \frac{11(x+z)(z+t)-(x+y+z+t)}{x+y+z+t +(x+z)(y+t)} $$
0 replies
SunnyEvan
42 minutes ago
0 replies
Simple inequality
sqing   60
N 42 minutes ago by Adywastaken
Source: Shortlist BMO 2018, A1
Let $a, b, c $ be positive real numbers such that $abc = \frac {2} {3}. $ Prove that:

$$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant  \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$$
60 replies
sqing
May 3, 2019
Adywastaken
42 minutes ago
Inequalities
sqing   21
N Today at 12:37 AM by sqing
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
21 replies
sqing
May 21, 2025
sqing
Today at 12:37 AM
Polar Coordinates
pingpongmerrily   4
N Today at 12:11 AM by K124659
Convert the equation $r=\tan(\theta)$ into rectangular form.
4 replies
pingpongmerrily
Today at 12:02 AM
K124659
Today at 12:11 AM
Calculus, sets
wl8418   2
N Yesterday at 11:18 PM by wl8418
Is empty set a proper set of an non empty set? Why or why not? Any clarification or insight is appreciated. Thanks in advance!
2 replies
wl8418
Yesterday at 6:11 AM
wl8418
Yesterday at 11:18 PM
Geometry
AlexCenteno2007   1
N Yesterday at 11:13 PM by ohiorizzler1434
Given triangle ABC, it is true that BD = CF where D and F are points in the same half-plane with respect to line BC and it is also known that BD is parallel to AC and CF is parallel to AB. Show that BF, CD and the interior bisector of A are concurrent.
1 reply
AlexCenteno2007
Yesterday at 10:02 PM
ohiorizzler1434
Yesterday at 11:13 PM
How do I prove this? - Sets and symmetric difference
smadadi1000   1
N Yesterday at 8:21 PM by KSH31415
For sets A, B and C, prove that (A Δ B) Δ C = (A Δ C) Δ (A \ B).

The textbook - Proofs by Jay Cummings - had this definition for symmetric difference:
A Δ B = ( A U B)/(A ∩ B)

this is exercise 3.37 (e).
1 reply
smadadi1000
Yesterday at 3:23 PM
KSH31415
Yesterday at 8:21 PM
help me ..
exoticc   2
N Yesterday at 5:00 PM by sodiumaka
Find all pairs of functions (f,g) : R->R that satisfy:
f(1)=2025;
g(f(x+y))+2x+y-1=f(x)+(2x+y)g(y) , ∀x,y∈R
2 replies
exoticc
Yesterday at 9:26 AM
sodiumaka
Yesterday at 5:00 PM
21st PMO National Orals #9
yes45   0
Yesterday at 3:22 PM
In square $ABCD$, $P$ and $Q$ are points on sides $CD$ and $BC$, respectively, such that $\angle{APQ} = 90^\circ$. If $AP = 4$ and $PQ = 3$, find the area of $ABCD$.

Answer Confirmation
Solution
0 replies
yes45
Yesterday at 3:22 PM
0 replies
Original Problem (qrxz17)
qrxz17   0
Yesterday at 1:32 PM
Problem. Suppose that
\begin{align*}
    (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 &= 138 \\
    (a^2+b^2+c^2)^2 &=100.
\end{align*}Find \(a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1)\).
Answer: Click to reveal hidden text
Solution. Subtracting the two given equations, we get
\begin{align*}
        a^4+b^4+c^4=38.
    \end{align*}Taking the square root of the second equation, we get

\begin{align*}
        a^2+b^2+c^2 = 10.
    \end{align*}
Then,
\begin{align*}
        a^4+b^4+c^4-(a^2+b^2+c^2) &= a^4-a^2+b^4-b^2+c^4-c^2 \\
        &= a^2(a^2-1)+b^2(b^2-1)+c^2(c^2-1) = \boxed{28}.
    \end{align*}
0 replies
qrxz17
Yesterday at 1:32 PM
0 replies
17th PMO Area Stage #5
yes45   0
Yesterday at 1:31 PM
Triangle \(ABC\) has a right angle at \(B\), with \(AB = 3\) and \(BC = 4\). If \(D\) and \(E\) are points
on \(AC\) and \(BC\), respectively, such that \(CD = DE = \frac{5}{3}\), find the perimeter of quadrilateral
\(ABED\).
Answer Confirmation
Solution
0 replies
yes45
Yesterday at 1:31 PM
0 replies
Sipnayan 2019 SHS Orals Semifinal Round B Difficult \#3
qrxz17   0
Yesterday at 1:30 PM
Problem. Suppose that \(a\), \(b\), \(c\) are positive integers such that \((a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 63\). Find all possible values of \(a+b+c\).
Answer. Click to reveal hidden text
Solution. Expanding and cancelling "like" terms, we get

\begin{align*}
        (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) &= (a^4 +b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2) - 2(a^4+b^4+c^4) \\
        &= 2a^2b^2+2a^2c^2+2b^2c^2 - a^4 -b^4-c^4
    \end{align*}
We multiply the entire equation by \(-1\) to rearrange the terms and place the higher-degree terms at the front of the expression in order to have a more recognizable form.

\begin{align*}
        a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 = -63.
    \end{align*}
Simplifying the left-hand expression, we get

\begin{align*}
        a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 &=a^4-2a^2(b^2+c^2)+b^4-2b^2c^2+c^4 \\
        &= a^4-2a^2(b^2+c^2)+(b^2+c^2)^2 - 4b^2c^2 \\
        &= [a^2 - (b^2+c^2)]^2-(2bc)^2 \\
        &= (a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc)\\
        &=[a^2-(b-c)^2][a^2-(b+c)^2] \\
        &= (a+b-c)(a-b+c)(a+b+c)(a-b-c) = -63.
    \end{align*}
If \(a=b=c\), we get
\begin{align*}
     -3a^4 &=-63 \\
     a^4 &= 21
    \end{align*}

in which there are no real solutions for \(a\).

If we either have \(a=b\), \(a=c\), or \(b=c\), we get something of the form

\[
    a^2(a+2b)(a-2b)=-63.
    \]
Since \(1\) and \(9\) are the only square factors of \(63\), we have \(a = \{1, 3\}\).

In this case, the solutions for \((a, b, c)\) are \((1, 4, 4)\) and \((3, 2, 2)\), including all their permutations.

If \(a, b, c\) are all distinct positive integers, there are no solutions.

Therefore, all possible values of \(a+b+c\) are \(\boxed{7 \text{ and }9}\).
0 replies
qrxz17
Yesterday at 1:30 PM
0 replies
area wanted, perpendiculars from vertices on int. and ext. angle bisectors
parmenides51   2
N Jan 21, 2021 by djmathman
Source: 2020-21 IOQM p22
In triangle $ABC$, let $P$ and $R$ be the feet of the perpendiculars from $A$ onto the external and internal bisectors of $\angle ABC$, respectively; and let $Q$ and $S$ be the feet of the perpendiculars from $A$ onto the internal and external bisectors of $\angle ACB$, respectively. If $PQ = 7, QR = 6$ and $RS = 8$, what is the area of triangle $ABC$?
2 replies
parmenides51
Jan 18, 2021
djmathman
Jan 21, 2021
area wanted, perpendiculars from vertices on int. and ext. angle bisectors
G H J
G H BBookmark kLocked kLocked NReply
Source: 2020-21 IOQM p22
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parmenides51
30653 posts
#1 • 1 Y
Y by Mango247
In triangle $ABC$, let $P$ and $R$ be the feet of the perpendiculars from $A$ onto the external and internal bisectors of $\angle ABC$, respectively; and let $Q$ and $S$ be the feet of the perpendiculars from $A$ onto the internal and external bisectors of $\angle ACB$, respectively. If $PQ = 7, QR = 6$ and $RS = 8$, what is the area of triangle $ABC$?
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KimSungOK
2 posts
#2
Y by
AB=13, AC = 14, BC = 15, S=84
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djmathman
7938 posts
#3
Y by
Of course, we can solve the problem more generally.

Observe that $P$, $Q$, $R$, and $S$ all lie on the $A$-midline of $\triangle ABC$, whence are collinear. Combining this with the fact that $APBR$ and $AQCS$ are rectangles yields $AB = PR = 13$, $AC = QS = 14$. Furthermore, if $Q'$ and $R'$ are the reflections of $A$ across $Q$ and $R$, respectively, then
\begin{align*}
BC &= BR' + CQ' - Q'R' \\
      &= BA + AC - 2RQ \\
      &= 13 + 14 - 2\cdot 6 = 15.
\end{align*}Hence area $84$.
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