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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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0 replies
jlacosta
May 1, 2025
0 replies
Inspired by Butterfly
sqing   1
N 12 minutes ago by sqing
Source: Own
Let $ a,b,c>0. $ Prove that
$$a^2+b^2+c^2+ab+bc+ca+abc-3(a+b+c) \geq 34-14\sqrt 7$$$$a^2+b^2+c^2+ab+bc+ca+abc-\frac{433}{125}(a+b+c) \geq \frac{2(57475-933\sqrt{4665})}{3125} $$
1 reply
+1 w
sqing
21 minutes ago
sqing
12 minutes ago
Prove the inequality
Butterfly   4
N 12 minutes ago by JARP091

Let $a,b,c,d$ be positive real numbers. Prove $$a^2+b^2+c^2+d^2+abcd-3(a+b+c+d)+7\ge 0.$$
4 replies
Butterfly
Yesterday at 12:36 PM
JARP091
12 minutes ago
3 var inequality
JARP091   0
18 minutes ago
Source: Own
Let \( x, y, z \in \mathbb{R}^+ \). Prove that
\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]without using the Rearrangement Inequality or Chebyshev's Inequality.
0 replies
2 viewing
JARP091
18 minutes ago
0 replies
Inequality
VicKmath7   17
N 24 minutes ago by math-olympiad-clown
Source: Balkan MO SL 2020 A2
Given are positive reals $a, b, c$, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that
$\frac{\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}}{3}\leq \frac{a+b+c-1}{\sqrt{2}}$.

Albania
17 replies
VicKmath7
Sep 9, 2021
math-olympiad-clown
24 minutes ago
Max and min of ab+bc+ca-abc
Tiira   6
N 5 hours ago by MathsII-enjoy
a, b and c are three non-negative reel numbers such that a+b+c=1.
What are the extremums of
ab+bc+ca-abc
?
6 replies
Tiira
Jan 29, 2021
MathsII-enjoy
5 hours ago
Polynomial Minimization
ReticulatedPython   4
N Today at 1:27 AM by jasperE3
Consider the polynomial $$p(x)=x^{n+1}-x^{n}$$, where $x, n \in \mathbb{R+}.$

(a) Prove that the minimum value of $p(x)$ always occurs at $x=\frac{n}{n+1}.$
4 replies
ReticulatedPython
May 6, 2025
jasperE3
Today at 1:27 AM
Trig Identity
gauss202   2
N Today at 1:24 AM by MathIQ.
Simplify $\dfrac{1-\cos \theta + \sin \theta}{\sqrt{1 - \cos \theta + \sin \theta - \sin \theta \cos \theta}}$
2 replies
gauss202
May 14, 2025
MathIQ.
Today at 1:24 AM
2018 Mock ARML I --7 2^n | \prod^{2048}_{k=0} C(2k , k)
parmenides51   3
N Today at 12:57 AM by MathIQ.
Find the largest integer $n$ such that $2^n$ divides $\prod^{2048}_{k=0} {2k \choose k}$.
3 replies
parmenides51
Jan 17, 2024
MathIQ.
Today at 12:57 AM
f(2x+y)=f(x+2y)
jasperE3   3
N Today at 12:46 AM by MathIQ.
Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that:
$$f(2x+y)=f(x+2y)$$for all $x,y>0$.
3 replies
jasperE3
Yesterday at 8:58 PM
MathIQ.
Today at 12:46 AM
System of Equations
P162008   1
N Yesterday at 8:31 PM by alexheinis
If $a,b$ and $c$ are real numbers such that

$(a + b)(b + c) = -1$

$(a - b)^2 + (a^2 - b^2)^2 = 85$

$(b - c)^2 + (b^2 - c^2)^2 = 75$

Compute $(a - c)^2 + (a^2 - c^2)^2.$
1 reply
P162008
Monday at 10:48 AM
alexheinis
Yesterday at 8:31 PM
Might be the first equation marathon
steven_zhang123   35
N Yesterday at 7:09 PM by lightsbug
As far as I know, it seems that no one on HSM has organized an equation marathon before. Click to reveal hidden textSo why not give it a try? Click to reveal hidden text Let's start one!
Some basic rules need to be clarified:
$\cdot$ If a problem has not been solved within $5$ days, then others are eligible to post a new probkem.
$\cdot$ Not only simple one-variable equations, but also systems of equations are allowed.
$\cdot$ The difficulty of these equations should be no less than that of typical quadratic one-variable equations. If the problem involves higher degrees or more variables, please ensure that the problem is solvable (i.e., has a definite solution, rather than an approximate one).
$\cdot$ Please indicate the domain of the solution to the equation (e.g., solve in $\mathbb{R}$, solve in $\mathbb{C}$).
Here's an simple yet fun problem, hope you enjoy it :P :
P1
35 replies
steven_zhang123
Jan 20, 2025
lightsbug
Yesterday at 7:09 PM
THREE People Meet at the SAME. TIME.
LilKirb   7
N Yesterday at 5:33 PM by hellohi321
Three people arrive at the same place independently, at a random between $8:00$ and $9:00.$ If each person remains there for $20$ minutes, what's the probability that all three people meet each other?

I'm already familiar with the variant where there are only two people, where you Click to reveal hidden text It was an AIME problem from the 90s I believe. However, I don't know how one could visualize this in a Click to reveal hidden text Help on what to do?
7 replies
LilKirb
Monday at 1:06 PM
hellohi321
Yesterday at 5:33 PM
Quite straightforward
steven_zhang123   1
N Yesterday at 3:16 PM by Mathzeus1024
Given that the sequence $\left \{ a_{n} \right \} $ is an arithmetic sequence, $a_{1}=1$, $a_{2}+a_{3}+\dots+a_{10}=144$. Let the general term $b_{n}$ of the sequence $\left \{ b_{n} \right \}$ be $\log_{a}{(1+\frac{1}{a_{n}} )} ( a > 0  \text{and}  a \ne  1)$, and let $S_{n}$ be the sum of the $n$ terms of the sequence $\left \{ b_{n} \right \}$. Compare the size of $S_{n}$ with $\frac{1}{3} \log_{a}{(1+\frac{1}{a_{n}} )} $.
1 reply
steven_zhang123
Jan 11, 2025
Mathzeus1024
Yesterday at 3:16 PM
Function and Quadratic equations help help help
Ocean_MathGod   1
N Yesterday at 11:26 AM by Mathzeus1024
Consider this parabola: y = x^2 + (2m + 1)x + m(m - 3) where m is constant and -1 ≤ m ≤ 4. A(-m-1, y1), B(m/2, y2), C(-m, y3) are three different points on the parabola. Now rotate the axis of symmetry of the parabola 90 degrees counterclockwise around the origin O to obtain line a. Draw a line from the vertex P of the parabola perpendicular to line a, meeting at point H.

1) express the vertex of the quadratic equation using an expression with m.
2) If, regardless of the value of m, the parabola and the line y=x−km (where k is a constant) have exactly one point of intersection, find the value of k.

3) (where I'm struggling the most) When 1 < PH ≤ 6, compare the values of y1, y2, and y3.
1 reply
Ocean_MathGod
Aug 26, 2024
Mathzeus1024
Yesterday at 11:26 AM
Two circles, a tangent line and a parallel
Valentin Vornicu   105
N May 15, 2025 by Fly_into_the_sky
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP = EQ$.
105 replies
Valentin Vornicu
Oct 24, 2005
Fly_into_the_sky
May 15, 2025
Two circles, a tangent line and a parallel
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
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MagicalToaster53
159 posts
#97 • 1 Y
Y by cubres
Notice that $MN$ bisects $\overline{AB},$ as the point where they intersect has equivalent powers to the respective circles. From this we discover that $M$ is the midpoint of $\overline{PQ}$ due to the homothety at $N$ mapping $\overline{PQ} \mapsto \overline{AB}.$ It then suffices to show $EM \perp PQ$.

Claim: $\triangle EAB \cong \triangle MAB.$
Proof: Observe that $AB$ bisects $\angle EAM:$ \[\angle EAB = \angle ECM = \angle ACM = \angle BAM. \]Similarly by symmetry, we find that $AB$ bisects $\angle EBM.$ Then by SAS, we conclude that $\triangle EAB \cong \triangle MAB,$ as was to be shown. $\square$

Now $EAMB$ is a kite, so that $EM \perp AB \implies EM \perp PQ,$ as $AB \parallel PQ$. Hence $\triangle EPQ$ is isosceles, as desired. $\blacksquare$
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lnzhonglp
120 posts
#98 • 1 Y
Y by cubres
Observe that $$\measuredangle EBA = \measuredangle EDC = \measuredangle ABM$$and $$\measuredangle EAB = \measuredangle ECM = \measuredangle MAB,$$so $\triangle EAB \cong \triangle MAB$ and $EM \perp AB$. From radical axis we see that $MN$ bisects $AB$, so by homothety at $N$ we find that $M$ is the midpoint of $PQ$. Then since $EM \perp PQ$ it follows that $EP = EQ$.
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Likeminded2017
391 posts
#99 • 1 Y
Y by cubres
Let $MN$ and $AB$ intersect at $K.$ As $K$ lies on the radical axis we have $KA=KB.$ Taking the homothety centered at $N$ that maps $M$ to $K$ we find that $MP=MQ.$ Then $\angle BAM=\angle MCA=\angle BAE$ and similarly $\angle EBA=\angle ABM$ so by ASA congruence $\triangle EAB \cong \triangle MAB.$ Thus $EM$ is perpendicular to $AB$ and thus $PQ$ so $EP=EQ.$
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ezpotd
1283 posts
#100 • 1 Y
Y by cubres
Letting $AB$ meet $MN$ at $K$, we see $KA^2 = KM \cdot KN = KB^2$, so $K$ is the midpoint of $AB$, taking the homothety from $AB$ to $PQ$ this gives $MP = MQ$, so it suffices to prove $EM \perp PQ$, or $EM \perp AB$. However, angle chasing gives $\angle EAB = \angle ECM = \angle AMC = \angle BAM, \angle EBA = \angle EDM = \angle BMD = \angle ABM$, so $\triangle EAB \cong \angle MAB$, so $E,M$ are reflections about $AB$, done.
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optimusprime154
23 posts
#101 • 2 Y
Y by SorPEEK, cubres
let \(Z \)= \(MN  \cap AB\) from PoP we know \(ZB^2 = ZM * ZN\) and \(ZA^2 = ZM * ZN\) which means \(ZA = ZB\) and since \(PQBA\) is a trapezoid, \(PM = MQ\)
now we prove \(EM \perp CD\) and we're done. label \(\angle ACM = x, \angle BDM = y\) we know from \(AB \parallel CD\) that \(\angle EAB = x\) and \(\angle EDC = y\)
we also know from \(AB\) being tangent that \(\angle MBA\ = y\) and \(\angle MAB = x\) the above facts indicate the similarity of \(\triangle MAB , \triangle EAB \) so \(EM \perp AB\) => \(EM \perp CD\) and we're done
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Maximilian113
575 posts
#102 • 1 Y
Y by cubres
Extend $MN$ to intersect $AB$ at $K,$ then clearly $AK=BK.$ Therefore $PM=MQ$ by the parallel lines.

Now, notice that $\angle EAB = \angle ACM = \angle BAM = \angle AMC.$ Therefore $AB$ bisects $\angle EAM$ and $AC=AM.$ Similarly $AB$ bisects $EBM$ and $BD=BM.$ Therefore, $\triangle EAB \cong \triangle MAB$ so $A, B$ are the circumcenters of $\triangle EMC, \triangle EMD$ respectively so by Thales' Theorem it follows that $\angle EMP = \angle EMQ = 90^\circ.$ Hence $EM$ is the perpendicular bisector of $PQ,$ and we are done. QED
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ehuseyinyigit
839 posts
#104 • 1 Y
Y by cubres
WELL-KNOWN. Since $AB\parallel CD$, we have $$\angle ACM=\angle AMC=\angle MAB=\angle BAE$$Thus, $AC=AM$ and $AB$ bisects $\angle EAM$. Similarly $BD=BM$ and $AB$ bisects $\angle EBM$ hence $ABME$ is a kite $\Rightarrow AE=AM=AC$ implying $EM\perp CD$. On the other hand, for point $F$ being intersection of lines $AB=MN$, $AF=BF$ implies $PM=MQ$. We obtain $EP=EQ$.

WELL-KNOWN. We will show $EN$ bisects $\angle CND$. This is true because $$\angle ENB=\angle EAB=\angle ENC$$and $\angle ANE=\angle ABE=\angle BND$. Thus, $EN$ bisects $\angle CND$.
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mathwiz_1207
103 posts
#105 • 1 Y
Y by cubres
We have
\[\measuredangle EAB = \measuredangle ACM = \measuredangle ANM = \measuredangle BAM = \measuredangle CMA\]\[\measuredangle EBA = \measuredangle BDM = \measuredangle BNM = \measuredangle ABM = \measuredangle DMB\], implying that $\triangle AMB \cong \triangle AEB$, and that $\triangle AMC, \triangle BDM$ are isosceles. Therefore,
\[AE = AM = AC\]so $A$ is the center of $(EMC)$, and similarly $B$ is the cetner of $(EMD)$. Thus, $EM \perp CD$. Now, let $L = MN \cap AB$, by Radical Axis,
\[LA = LB\]and since $\triangle NPQ \sim \triangle NAB$, we have $MP = MQ \implies EQ = EP$, so we are done.
This post has been edited 1 time. Last edited by mathwiz_1207, Feb 23, 2025, 11:53 PM
Reason: forgot to add last line
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study1126
569 posts
#106 • 1 Y
Y by cubres
All angles will be directed in this solution.

First, $\angle EAB=\angle ECB=\angle BAM$, since $AB||CD$ and $AB$ is tangent to the two circles. Similarly, $\angle EBA=\angle ABM$. Combined with $AB=AB$ gives $\triangle EAB\cong \triangle MAB$.

This implies $EAMB$ is a kite, so $EM\perp AB$, which means $EM\perp CD$. Since $M$ is on the radical axis, extending $NM$ to $X$ on $AB$ gives $X$ is the midpoint of $AB$. Homothety shows $M$ is the midpoint of $PQ$. Thus, $MP=MQ$, so by congruent triangle $\triangle EPM\cong \triangle EQM$, $EP=EQ$, as desired.
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joshualiu315
2534 posts
#107 • 1 Y
Y by cubres
It is well-known that $\overline{MN}$ bisects $\overline{AB}$, and since $\overline{AB} \parallel \overline{CD}$, we also know that $\overline{MN}$ bisects $\overline{PQ}$. This implies $M$ is the midpoint of $\overline{PQ}$.

We have $\angle EAB = \angle ACM = \angle MAB$ and similarly, $\angle EBA = \angle BDM = \angle MBA$. This means $\triangle EAB \cong \triangle MAB$, which implies that $\overline{EM} \perp \overline{AB}$. Because $\overline{AB} \parallel \overline{CD}$, we have $\overline{EM} \perp \overline{PQ}$, so $\overline{EM}$ is the perpendicular bisector of $\overline{PQ}$. Thus, $EP=EQ$, as desired. $\blacksquare$
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LeYohan
53 posts
#108 • 1 Y
Y by cubres
First IMO #1 solve and it's geo!! :D

Claim: $PM = MQ$.
Let $NM$ intersect $AB$ at $X$. It's well known that the radical axis intersects the common tangent, hence $AX = XB$. By the Thales Theorem, we inmediately get the result. $\square$

Now doing some angle chase using the fact that $CD \parallel AB$ and $AB$ is tangent to both circles, we get:
$\angle EBA = \angle BDM = \angle BNM = \angle ABM = \angle BMD$, and $\angle EAB = \angle ACM = \angle ANM = \angle BAM = \angle AMC$. As a result, $\triangle ACM$ and $\triangle BMC$ are isosceles. Furthermore, $\triangle MAB \cong \triangle EAB \implies EB = BD, EA = AC$. This means that $\angle EMC = \angle EMD = 90$, and notice that we originally proved that $PM = MQ$, so $\triangle EPQ$ is isosceles and we're done. $\square$
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Avron
37 posts
#109 • 1 Y
Y by cubres
$MN$ is the radical axis of $G_1,G_2$ so it bisects $AB$ so we also know that $M$ is the midpoint of $PQ$, thus it is enough to prove $ME\perp CD$.
Now notice that $\angle MAB=\angle ACM = \angle EAB$ and similarly $\angle MBA = \angle ABE$ so $EAB \cong MAB$ and we're done
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zuat.e
66 posts
#110 • 1 Y
Y by cubres
Easy angle chasing yields the cyclicity of $(BEAN)$. We claim the following:

Claim: $EA=AM=AC$ and similarly $EB=BM=BD$
Proof: Let $O_1$ be the center of $G_1$. It is clear that $O_1A$ is the side bisector of $MC$, hence $AM=AC$.
Furthermore, note that $\measuredangle BAE=\measuredangle DCE= \measuredangle BMC=\measuredangle MAB$ and similarly $\measuredangle EBA =\measuredangle ABM$, hence $\triangle EAB\cong MAB$ (because they also share $AB$).

It now follows that $AC=AM=AE$ and analogously $BD=BM=BE$
Finally, as $\measuredangle MCA= \measuredangle MNA=\measuredangle ANC$, we have $AP\cdot AN=AC^2=AE^2$, hence $\measuredangle CEP =\measuredangle ENA=\measuredangle EBA=\measuredangle EDC$ and similarly $\measuredangle QED=\measuredangle PCE$, yielding $\measuredangle CPE =\measuredangle EQD$, as desired.
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cubres
119 posts
#111
Y by
ACGN
MOHS Hardness Scale
IMO Problem #1 (May 2) - IMO 2001 p1
This post has been edited 1 time. Last edited by cubres, May 7, 2025, 8:15 PM
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Fly_into_the_sky
3 posts
#112
Y by
Smooth one :pilot:
Claim 1: $EM \perp PQ$
Proof: Notice that $\angle{EAB}=\angle{ACM}=\angle{MAB}$ and $\angle{EBA}=\angle{BDM}=\angle{MBA}$ and $AB$ is common side
So $\triangle AEB \cong \triangle AMB \implies EM \perp AB \implies EM \perp PQ$
Claim 2: $MP=MQ$
Proof: Let $R=(MN) \cap (AB)$ then by PoP, $R$ is the midpoint of $AB$ and so $M$ is the midpoint of $(PQ)$ by playing with Thales theorem on $\triangle RNB$ and $\triangle RAN$
Conclude
This post has been edited 2 times. Last edited by Fly_into_the_sky, May 15, 2025, 5:26 PM
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