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Thailand geometry
EeEeRUT   3
N 5 minutes ago by ItzsleepyXD
Source: Thailand MO 2025 P7
Let $ABC$ be a triangle with $AB < AC$. The tangent to the circumcircle of $\triangle ABC$ at $A$ intersects $BC$ at $D$. The angle bisector of $\angle BAC$ intersect $BC$ at $E$. Suppose that the perpendicular bisector of $AE$ intersect $AB, AC$ at $P,Q$, respectively. Show that $$\sqrt{\frac{BP}{CQ}} = \frac{AC \cdot BD}{AB \cdot CD}$$
3 replies
EeEeRUT
2 hours ago
ItzsleepyXD
5 minutes ago
Collinearity of intersection points in a triangle
MathMystic33   2
N 26 minutes ago by AylyGayypow009
Source: 2025 Macedonian Team Selection Test P1
On the sides of the triangle \(\triangle ABC\) lie the following points: \(K\) and \(L\) on \(AB\), \(M\) on \(BC\), and \(N\) on \(CA\). Let
\[
P = AM\cap BN,\quad
R = KM\cap LN,\quad
S = KN\cap LM,
\]and let the line \(CS\) meet \(AB\) at \(Q\). Prove that the points \(P\), \(Q\), and \(R\) are collinear.
2 replies
MathMystic33
Yesterday at 5:56 PM
AylyGayypow009
26 minutes ago
Parallel lines in incircle configuration
GeorgeRP   1
N 40 minutes ago by Double07
Source: Bulgaria IMO TST 2025 P1
Let $I$ be the incenter of triangle $\triangle ABC$. Let $H_A$, $H_B$, and $H_C$ be the orthocenters of triangles $\triangle BCI$, $\triangle ACI$, and $\triangle ABI$, respectively. Prove that the lines through $H_A$, $H_B$, and $H_C$, parallel to $AI$, $BI$, and $CI$, respectively, are concurrent.
1 reply
1 viewing
GeorgeRP
2 hours ago
Double07
40 minutes ago
Proving that these are concyclic.
Acrylic3491   0
an hour ago
In $\bigtriangleup ABC$, points $P$ and $Q$ are isogonal conjugates. The tangent to $(BPC)$ at $P$ and the tangent to $(BQC)$ at Q, meet at $R$. $AR$ intersects $(ABC)$ at $D$. Prove that points $P$,$Q$, $R$ and $D$ are concyclic.

Any hints on this ?
0 replies
Acrylic3491
an hour ago
0 replies
Inspired by old results
sqing   4
N an hour ago by sqing
Source: Own
Let $ a,b,c>0 $ . Prove that
$$\frac{a+kb}{b+c}+\frac{b+kc}{c+a}+\frac{c+ka}{a+b}\geq \frac{3(k+1)}{2}$$W here $-1 \leq k \leq  \frac{537}{90}.$
4 replies
1 viewing
sqing
Today at 3:58 AM
sqing
an hour ago
BMO Shortlist 2021 G2
Lukaluce   6
N an hour ago by tilya_TASh
Source: BMO Shortlist 2021
Let $I$ and $O$ be the incenter and the circumcenter of a triangle $ABC$, respectively, and let
$s_a$ be the exterior bisector of angle $\angle BAC$. The line through $I$ perpendicular to $IO$ meets the
lines $BC$ and $s_a$ at points $P$ and $Q$, respectively. Prove that $IQ = 2IP$.
6 replies
Lukaluce
May 8, 2022
tilya_TASh
an hour ago
Interesting inequalities
sqing   2
N an hour ago by sqing
Source: Own
Let $a,b,c >1 $ and $ \frac{1}{a-1}+\frac{1}{b-1}+\frac{1}{c-1}=1.$ Show that$$  \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}   \leq \frac{3}{5}$$$$ \frac{1}{a+1}+\frac{1}{b+2}+\frac{1}{c+1} \leq \frac{4}{7}$$$$  \frac{1}{a+2}+\frac{1}{b+1}+\frac{1}{c+2}  \leq \frac{7}{13}$$
2 replies
sqing
Yesterday at 11:48 AM
sqing
an hour ago
ISI UGB 2025 P2
SomeonecoolLovesMaths   9
N 2 hours ago by SatisfiedMagma
Source: ISI UGB 2025 P2
If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^2 A + \sin ^2 B + \sin^2  C = 2 \left( \cos ^2 A + \cos ^2 B + \cos ^2 C \right),$$prove that the triangle must have a right angle.
9 replies
SomeonecoolLovesMaths
May 11, 2025
SatisfiedMagma
2 hours ago
inequality
xytunghoanh   6
N 2 hours ago by sqing
For $a,b,c\ge 0$. Let $a+b+c=3$.
Prove or disprove
\[\sum ab +\sum ab^2 \le 6\]
6 replies
xytunghoanh
6 hours ago
sqing
2 hours ago
Number theory
EeEeRUT   1
N 2 hours ago by lgx57
Source: Thailand MO 2025 P10
Let $n$ be a positive integer. Show that there exist a polynomial $P(x)$ with integer coefficient that satisfy the following
[list]
[*]Degree of $P(x)$ is at most $2^n - n -1$
[*]$|P(k)| = (k-1)!(2^n-k)!$ for each $k \in \{1,2,3,\dots,2^n\}$
[/list]
1 reply
EeEeRUT
3 hours ago
lgx57
2 hours ago
Trigonometric Product
Henryfamz   1
N 2 hours ago by pco
Compute $$\prod_{n=1}^{45}\sin(2n-1)$$
1 reply
Henryfamz
Yesterday at 4:52 PM
pco
2 hours ago
Ratio of lengths in right-angled triangle
DylanN   2
N Apr 8, 2025 by sunken rock
Source: South African Mathematics Olympiad 2021, Problem 2
Let $PAB$ and $PBC$ be two similar right-angled triangles (in the same plane) with $\angle PAB = \angle PBC = 90^\circ$ such that $A$ and $C$ lie on opposite sides of the line $PB$. If $PC = AC$, calculate the ratio $\frac{PA}{AB}$.
2 replies
DylanN
Aug 11, 2021
sunken rock
Apr 8, 2025
Ratio of lengths in right-angled triangle
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Source: South African Mathematics Olympiad 2021, Problem 2
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DylanN
194 posts
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Let $PAB$ and $PBC$ be two similar right-angled triangles (in the same plane) with $\angle PAB = \angle PBC = 90^\circ$ such that $A$ and $C$ lie on opposite sides of the line $PB$. If $PC = AC$, calculate the ratio $\frac{PA}{AB}$.
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Mathzeus1024
886 posts
#2
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Let $a,b>0$ such that $P(a,b); A(0,b); B(0,0)$ exist in the first quadrant of the $xy-$plane for right $\Delta PAB$. If hypothenuse $BP$ is represented by the line $y=\frac{b}{a}x$, then let side $BC$ of similar right $\Delta PBC$ be represented by $y = -\frac{a}{b}x$ such that $C\left(t, -\frac{at}{b}\right)$. We are given that $|PC|=|AC|$, which satisfies:

$\sqrt{(a-t)^2+\left(b+\frac{at}{b}\right)^2} = \sqrt{t^2 + \left(b+\frac{at}{b}\right)^2} \Rightarrow t =\frac{a}{2}$

or $C\left(\frac{a}{2}, -\frac{a^2}{2b}\right)$. Next, let:

$|PB|=k|PA| \Rightarrow \sqrt{a^2+b^2}= ka$ (i);

$|PC|=k|PB| \Rightarrow \sqrt{\left(a-\frac{a}{2}\right)^2+\left(b+\frac{a^2}{2b}\right)^2} = k\sqrt{a^2+b^2}$ (ii)

of which we obtain:

$\frac{a^2+b^2}{a} = \sqrt{\left(a-\frac{a}{2}\right)^2+\left(b+\frac{a^2}{2b}\right)^2}$;

or $\frac{(a^2+b^2)^2}{a^2} = \left(\frac{a}{2}\right)^2 + \left(\frac{2b^2+a^2}{2b}\right)^2$;

or $4a^4b^2+8a^2b^4+4b^6 = a^4b^2+4a^2b^4+4a^4b^2+a^6$;

or $0 = (a^2+b^2)(a^2+2b^2)(a+\sqrt{2}b)(a-\sqrt{2}b) \Rightarrow a=\sqrt{2}b$ (iii).

Hence, $\frac{|PA|}{|AB|} = \frac{a}{b} = \textcolor{red}{\sqrt{2}}$.
This post has been edited 7 times. Last edited by Mathzeus1024, Apr 7, 2025, 12:38 PM
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sunken rock
4394 posts
#3 • 1 Y
Y by Mathzeus1024
My solution with less calculations at https://stanfulger.blogspot.com/2025/04/aops-httpsartofproblemsolvingcomcommuni.html.

Best regards,
sunken rock
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