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Prove excircle is tangent to circumcircle
sarjinius   7
N 13 minutes ago by markam
Source: Philippine Mathematical Olympiad 2025 P4
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.
7 replies
sarjinius
Mar 9, 2025
markam
13 minutes ago
An easy FE
oVlad   1
N 36 minutes ago by pco
Source: Romania EGMO TST 2017 Day 1 P3
Determine all functions $f:\mathbb R\to\mathbb R$ such that \[f(xy-1)+f(x)f(y)=2xy-1,\]for any real numbers $x{}$ and $y{}.$
1 reply
oVlad
3 hours ago
pco
36 minutes ago
Fractions and reciprocals
adihaya   34
N 36 minutes ago by de-Kirschbaum
Source: 2013 BAMO-8 #4
For a positive integer $n>2$, consider the $n-1$ fractions $$\dfrac21, \dfrac32, \cdots, \dfrac{n}{n-1}$$The product of these fractions equals $n$, but if you reciprocate (i.e. turn upside down) some of the fractions, the product will change. Can you make the product equal 1? Find all values of $n$ for which this is possible and prove that you have found them all.
34 replies
adihaya
Feb 27, 2016
de-Kirschbaum
36 minutes ago
GCD Functional Equation
pinetree1   60
N 38 minutes ago by cursed_tangent1434
Source: USA TSTST 2019 Problem 7
Let $f: \mathbb Z\to \{1, 2, \dots, 10^{100}\}$ be a function satisfying
$$\gcd(f(x), f(y)) = \gcd(f(x), x-y)$$for all integers $x$ and $y$. Show that there exist positive integers $m$ and $n$ such that $f(x) = \gcd(m+x, n)$ for all integers $x$.

Ankan Bhattacharya
60 replies
pinetree1
Jun 25, 2019
cursed_tangent1434
38 minutes ago
Inequality
giangtruong13   3
N 43 minutes ago by KhuongTrang
Let $a,b,c >0$ such that: $a^2+b^2+c^2=3$. Prove that: $$\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+abc \geq 4$$
3 replies
giangtruong13
Today at 8:01 AM
KhuongTrang
43 minutes ago
Easy geo
oVlad   3
N an hour ago by Primeniyazidayi
Source: Romania EGMO TST 2019 Day 1 P1
A line through the vertex $A{}$ of the triangle $ABC{}$ which doesn't coincide with $AB{}$ or $AC{}$ intersectes the altitudes from $B{}$ and $C{}$ at $D{}$ and $E{}$ respectively. Let $F{}$ be the reflection of $D{}$ in $AB{}$ and $G{}$ be the reflection of $E{}$ in $AC{}.$ Prove that the circles $ABF{}$ and $ACG{}$ are tangent.
3 replies
oVlad
3 hours ago
Primeniyazidayi
an hour ago
abc(a+b+c)=3, show that prod(a+b)>=8 [Indian RMO 2012(b) Q4]
Potla   28
N an hour ago by mihaig
Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that we have
\[(a+b)(b+c)(c+a)\geq 8.\]
Also determine the case of equality.
28 replies
Potla
Dec 2, 2012
mihaig
an hour ago
NT with repeating decimal digits
oVlad   1
N an hour ago by kokcio
Source: Romania EGMO TST 2019 Day 1 P2
Determine the digits $0\leqslant c\leqslant 9$ such that for any positive integer $k{}$ there exists a positive integer $n$ such that the last $k{}$ digits of $n^9$ are equal to $c{}.$
1 reply
oVlad
3 hours ago
kokcio
an hour ago
Inequalities make a comeback
MS_Kekas   2
N an hour ago by ZeroHero
Source: Kyiv City MO 2025 Round 1, Problem 11.5
Determine the largest possible constant \( C \) such that for any positive real numbers \( x, y, z \), which are the sides of a triangle, the following inequality holds:
\[
\frac{xy}{x^2 + y^2 + xz} + \frac{yz}{y^2 + z^2 + yx} + \frac{zx}{z^2 + x^2 + zy} \geq C.
\]
Proposed by Vadym Solomka
2 replies
MS_Kekas
Jan 20, 2025
ZeroHero
an hour ago
Interesting F.E
Jackson0423   11
N an hour ago by Jackson0423
Show that there does not exist a function
\[
f : \mathbb{R}^+ \to \mathbb{R}
\]satisfying the condition that for all \( x, y \in \mathbb{R}^+ \),
\[
f(x + y^2) \geq f(x) + y.
\]

~Korea 2017 P7
11 replies
Jackson0423
Apr 18, 2025
Jackson0423
an hour ago
Sequence...
Jackson0423   0
an hour ago
Let the sequence \( \{a_n\} \) be defined as follows:
\( a_0 = 1 \), and for all positive integers \( n \),
\[
a_n = a_{\left\lfloor \frac{n}{3} \right\rfloor} + a_{\left\lfloor \frac{n}{2} \right\rfloor}.
\]Find the sum of all values \( k \leq 100 \) for which there exists a unique positive integer \( n \) such that \( a_n = k \).
0 replies
Jackson0423
an hour ago
0 replies
Geometry problem
kjhgyuio   1
N Apr 4, 2025 by Mathzeus1024
Source: smo
In trapezium ABCD,AD is parallel to BC and points E and F are midpoints of AB and DC respectively. If
Area of AEFD/Area of EBCF =√3 + 1/3-√3 and the area of triangle ABD is √3 .find the area of trapezium ABCD
1 reply
kjhgyuio
Apr 1, 2025
Mathzeus1024
Apr 4, 2025
Geometry problem
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Source: smo
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kjhgyuio
49 posts
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In trapezium ABCD,AD is parallel to BC and points E and F are midpoints of AB and DC respectively. If
Area of AEFD/Area of EBCF =√3 + 1/3-√3 and the area of triangle ABD is √3 .find the area of trapezium ABCD
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Mathzeus1024
823 posts
#2
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In the first quadrant of the $xy-$plane, let:

$A(0,0); B(h\cot(A),h); C(k-h\cot(D),h); D(k,0); E\left(\frac{h\cot(A)}{2},\frac{h}{2}\right); F\left(k - \frac{h\cot(D)}{2}, \frac{h}{2}\right)$

define trapezoid $ABCD$ that is divided into trapezoids $AEFD$ and $EBCF$ by segment $EF$ (for $AD \parallel EF \parallel BC$).

We have $[ABD] = \frac{kh}{2}=\sqrt{3}$. If $[AEFD] = \left(\sqrt{3}+\frac{1}{3-\sqrt{3}}\right)[EBCF]$, then we obtain:

$\frac{h/2}{2}\left[k + \left(k-\frac{h\cot(A)}{2}-\frac{h\cot(D)}{2}\right)\right] = \left(\sqrt{3}+\frac{1}{3-\sqrt{3}}\right) \cdot \frac{h/2}{2}\left[\left(k-\frac{h\cot(A)}{2}-\frac{h\cot(D)}{2}\right) + (k-h\cot(A)-h\cot(D))\right]$;

or $2k + -\frac{h\cot(A)}{2}-\frac{h\cot(D)}{2} = \left(\sqrt{3}+\frac{1}{3-\sqrt{3}}\right) \cdot \left[2k-\frac{3h\cot(A)}{2}-\frac{3h\cot(D)}{2}\right]$;

or $\frac{4\sqrt{3}}{h}  -\frac{h\cot(A)}{2}-\frac{h\cot(D)}{2} = \left(\sqrt{3}+\frac{1}{3-\sqrt{3}}\right) \cdot \left[\frac{4\sqrt{3}}{h}-\frac{3h\cot(A)}{2}-\frac{3h\cot(D)}{2}\right]$;

or $8\sqrt{3} - h^{2}[\cot(A)+\cot(D)] = \left(\sqrt{3}+\frac{1}{3-\sqrt{3}}\right) \cdot [8\sqrt{3} - 3h^{2}[\cot(A)+\cot(D)]$;

or $[\cot(A)+\cot(D]h^2 = \frac{56-8\sqrt{3}}{1+7\sqrt{3}}$ (i).

Now, $[ABCD] = \frac{h}{2}[2k-h\cot(A)-h\cot(D)] = hk-\frac{[\cot(A)+\cot(D]h^2}{2}$;

or $2\sqrt{3} - \frac{1}{2} \cdot \frac{56-8\sqrt{3}}{1+7\sqrt{3}} = 2\sqrt{3} - \frac{28-4\sqrt{3}}{1+7\sqrt{3}} = \textcolor{red}{\frac{56+46\sqrt{3}}{73}}$.
This post has been edited 4 times. Last edited by Mathzeus1024, Apr 4, 2025, 9:22 AM
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