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Symmedian
Pomegranat   1
N 41 minutes ago by xytunghoanh
Source: Russian geometry olympiad
In triangle $ABC$, point $M$ is the midpoint of $BC$, $P$ the point of intersection of the tangents at points $B$ and $C$ of the circumscribed circle of $ABC$, $N$ is the midpoint of the segment $MP$. The segment $AN$ meets the circumcircle $ABC$ at the point $Q$. Prove that $\angle PMQ = \angle MAQ$.
1 reply
Pomegranat
an hour ago
xytunghoanh
41 minutes ago
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P>2D
gwen01   3
N an hour ago by AshAuktober
Source: Baltic Way 1992 #18
Show that in a non-obtuse triangle the perimeter of the triangle is always greater than two times the diameter of the circumcircle.
3 replies
gwen01
Feb 18, 2009
AshAuktober
an hour ago
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inequality with interesting conditions
Cobedangiu   2
N an hour ago by musava_ribica
Let $x,y,z>0$:
2 replies
Cobedangiu
an hour ago
musava_ribica
an hour ago
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Hard geometry proof
radhoan_rikto-   1
N an hour ago by GreekIdiot
Source: BDMO 2025
Let ABC be an acute triangle and D the foot of the altitude from A onto BC. A semicircle with diameter BC intersects segments AB,AC and AD in the points F,E and X respectively.The circumcircles of the triangles DEX and DFX intersect BC in L and N respectively, other than D. Prove that BN=LC.
1 reply
radhoan_rikto-
Apr 25, 2025
GreekIdiot
an hour ago
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Inspired by JK1603JK
sqing   0
2 hours ago
Source: Own
Let $ a,b,c $ be reals such that $ abc\neq 0$ and $ a+b+c=0. $ Prove that
$$\left|\frac{a-b}{c}\right|+k\left|\frac{b-c}{a} \right|+k^2\left|\frac{c-a}{b} \right|\ge 3(k+1)$$Where $ k>0.$
$$\left|\frac{a-b}{c}\right|+2\left|\frac{b-c}{a} \right|+4\left|\frac{c-a}{b} \right|\ge 9$$
0 replies
sqing
2 hours ago
0 replies
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problem interesting
Cobedangiu   9
N 2 hours ago by Cobedangiu
Let $a=3k^2+3k+1 (a,k \in N)$
$i)$ Prove that: $a^2$ is the sum of $3$ square numbers
$ii)$ Let $b \vdots a$ and $b$ is the sum of $3$ square numbers. Prove that: $b^n$ is the sum of $3$ square numbers
9 replies
Cobedangiu
Yesterday at 5:06 AM
Cobedangiu
2 hours ago
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4-var inequality
RainbowNeos   0
2 hours ago
Given $a,b,c,d>0$, show that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]
0 replies
RainbowNeos
2 hours ago
0 replies
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Find all integer pairs (m,n) such that 2^n! + 1 | 2^m! + 19
Goblik   0
2 hours ago
Find all positive integer pairs $(m,n)$ such that $2^{n!} + 1 | 2^{m!} + 19$
0 replies
Goblik
2 hours ago
0 replies
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Junior Balkan Mathematical Olympiad 2024- P3
Lukaluce   15
N 2 hours ago by MATHS_ENTUSIAST
Source: JBMO 2024
Find all triples of positive integers $(x, y, z)$ that satisfy the equation

$$2020^x + 2^y = 2024^z.$$
Proposed by Ognjen Tešić, Serbia
15 replies
Lukaluce
Jun 27, 2024
MATHS_ENTUSIAST
2 hours ago
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AD is Euler line of triangle IKL
VicKmath7   16
N 2 hours ago by ErTeeEs06
Source: IGO 2021 Advanced P5
Given a triangle $ABC$ with incenter $I$. The incircle of triangle $ABC$ is tangent to $BC$ at $D$. Let $P$ and $Q$ be points on the side BC such that $\angle PAB = \angle BCA$ and $\angle QAC = \angle ABC$, respectively. Let $K$ and $L$ be the incenter of triangles $ABP$ and $ACQ$, respectively. Prove that $AD$ is the Euler line of triangle $IKL$.

Proposed by Le Viet An, Vietnam
16 replies
VicKmath7
Dec 30, 2021
ErTeeEs06
2 hours ago
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Putnam 1958 February A5
sqrtX   3
N Today at 2:19 AM by centslordm
Source: Putnam 1958 February
Show that the integral equation
$$f(x,y) = 1 + \int_{0}^{x} \int_{0}^{y} f(u,v) \, du \, dv$$has at most one solution continuous for $0\leq x \leq 1, 0\leq y \leq 1.$
3 replies
sqrtX
Jul 18, 2022
centslordm
Today at 2:19 AM
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integral
Svyatoslav   5
N Nov 16, 2024 by tobiSALT
How do you find
$$I(a,b)=\int_0^\infty\frac{\ln|b-\cos ax|}{1+x^2}\,dx,\,\,b\in[0;1]\,\,?$$
5 replies
Svyatoslav
Nov 14, 2024
tobiSALT
Nov 16, 2024
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integral
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Reveal topic tags
integration
calculus
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Svyatoslav
540 posts
Svyatoslav
#1 h  Nov 14, 2024, 6:05 AM • 1 Y
Y by Martin.s
How do you find
$$I(a,b)=\int_0^\infty\frac{\ln|b-\cos ax|}{1+x^2}\,dx,\,\,b\in[0;1]\,\,?$$
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tobiSALT
81 posts
tobiSALT
#2 h  Nov 15, 2024, 6:43 AM • 1 Y
Y by StarLex1
Let the integral be
$ I(a, b) = \int_0^{\infty} \frac{\ln |b - \cos(ax)|}{1 + x^2} \, dx, \quad b \in [0, 1] \text{ and } a > 0. $
Let $b = \cos \theta$ for some $\theta \in [0, \pi/2]$. Then we can rewrite the integral as
$ I(a, b) = \int_0^{\infty} \frac{\ln |\cos \theta - \cos(ax)|}{1 + x^2} \, dx. $
We have the identity $\cos \theta - \cos(ax) = 2 \sin \frac{ax + \theta}{2} \sin \frac{ax - \theta}{2}$. Thus,
$ \ln |\cos \theta - \cos(ax)| = \ln |2 \sin \frac{ax+\theta}{2} \sin \frac{ax-\theta}{2}| = \ln 2 + \ln |\sin \frac{ax+\theta}{2}| + \ln |\sin \frac{ax-\theta}{2}|. $
Also, we know that
$ \int_0^{\infty} \frac{\ln 2}{1 + x^2} \, dx = \ln 2 \int_0^{\infty} \frac{dx}{1 + x^2} = \ln 2 \arctan(x) \Big|_0^{\infty} = \ln 2 \cdot \frac{\pi}{2} = \frac{\pi}{2} \ln 2. $
We use the known result
$ \int_0^{\infty} \frac{\ln|\sin(rx)|}{1 + x^2} \, dx = \frac{\pi}{2} \ln \left( \frac{1 - e^{-2r}}{2} \right). $
Then
$ \int_0^{\infty} \frac{\ln |\sin \frac{ax+\theta}{2}|}{1 + x^2} \, dx = \int_0^{\infty} \frac{\ln |\sin \frac{a}{2} (x+\theta/a)|}{1 + x^2} \, dx $
and
$ \int_0^{\infty} \frac{\ln |\sin \frac{ax-\theta}{2}|}{1 + x^2} \, dx = \int_0^{\infty} \frac{\ln |\sin \frac{a}{2} (x-\theta/a)|}{1 + x^2} \, dx. $
So we have
$ I(a, b) = \int_0^{\infty} \frac{\ln |b - \cos(ax)|}{1 + x^2} \, dx = \int_0^{\infty} \frac{\ln |\cos \theta - \cos(ax)|}{1 + x^2} \, dx = \frac{\pi}{2} \ln 2 + \int_0^{\infty} \frac{\ln |\sin(\frac{ax+\theta}{2})| + \ln |\sin(\frac{ax-\theta}{2})|}{1 + x^2} \, dx. $
If $\theta = 0$, i.e., $b = 1$, then
$ \int_0^{\infty} \frac{\ln(1 - \cos(ax))}{1 + x^2} \, dx = \int_0^{\infty} \frac{2 \ln |\sin(ax/2)| + \ln 2}{1 + x^2} \, dx = \frac{\pi}{2} \ln 2 + 2 \int_0^{\infty} \frac{\ln |\sin(ax/2)|}{1 + x^2} \, dx. $
Using the formula above, we get
$ \frac{\pi}{2} \ln 2 + 2 \frac{\pi}{2} \ln \left( \frac{1 - e^{-a}}{2} \right) = \frac{\pi}{2} (\ln 2 + 2 \ln (1 - e^{-a}) - 2 \ln 2) = \pi \ln(1 - e^{-a}) - \frac{\pi}{2} \ln 2. $
Thus
$ I(a, 1) = \pi \ln(1 - e^{-a}) - \frac{\pi}{2} \ln 2. $
Therefore, our final answer is
$ I(a, b) = \pi \ln \left( \frac{1 + \sqrt{1-b^2} - e^{-a}}{2} \right). $
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Svyatoslav
540 posts
Svyatoslav
#3 h  Nov 15, 2024, 12:54 PM • 1 Y
Y by tobiSALT
Thank you for your solution, but I think the answer is not correct. At $b=0\,$ and $\,a=0$
$$\int_0^\infty\frac{\ln (1)}{1+x^2}dx=0\neq\,\pi\ln\frac12$$There is also a known result (I.S. Gradshteyn and I.M. Ryzhik, GW (338)(28a), p 569)
$$\int_0^\infty\frac{\ln\cos^2ax}{1+x^2}dx=\pi\ln\frac{1+e^{-2a}}2$$
This post has been edited 1 time. Last edited by Svyatoslav, Nov 15, 2024, 6:42 PM
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tobiSALT
81 posts
tobiSALT
#4 h  Nov 15, 2024, 1:08 PM
Y by
True. I´ll try again later today, this integral is really hard.
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Svyatoslav
540 posts
Svyatoslav
#5 h  Nov 15, 2024, 1:20 PM • 1 Y
Y by Martin.s
Thank you. I got
$$I(a,b)=\frac\pi2\ln\frac{1-2be^{-a}+e^{-2a}}2, \,b\in [-1;1];\,a\geqslant0$$what gives the correct answer for one of the options: $b=0$ or $b=1$ or $a=0$. My problem is that WA (free option) is not able to evaluate the integral numerically, and I cannot get a reliable numeric check.

PS The answer is indeed correct - please see the solution.
This post has been edited 2 times. Last edited by Svyatoslav, Nov 16, 2024, 1:18 AM
Reason: slight correction
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tobiSALT
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tobiSALT
#7 h  Nov 16, 2024, 5:07 PM • 1 Y
Y by Svyatoslav
Great! Good solution
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