New geometry problem

$\bullet$ Write only the answers to the questions given.
$\bullet$ Some questions can have more than one correct answer. You are asked to provide the most correct or exact answer to a question like this. Scores will only be given to the giver of the most correct or most exact answer.
$\bullet$ Each question is worth 1 (one) point.
$\bullet \bullet$ to be more exact:
$\rhd$ in years 2002-08 time was 90' for part A and 120' for part B
$\rhd$ since years 2009 time is 210' for part A and B totally
$\rhd$ each problem in part A is 1 point, in part B is 7 points

p1. In the bag there are $7$ red balls and $8$ white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ...

p2. Given a regular hexagon with a side length of $1$ unit. The area of the hexagon is ...

p3. It is known that $r, s$ and $1$ are the roots of the cubic equation $x^3 - 2x + c = 0$ . The value of $(r-s)^2$ is ...

p4. The number of pairs of natural numbers $(m, n)$ so that $GCD(n,m) = 2$ and $LCM(m,n) = 1000$ is ...

p5. A data with four real numbers $2n-4$ , $2n-6$ , $n^2-8$ , $3n^2-6$ has an average of $0$ and a median of $9/2$ . The largest number of such data is ...

p6. Suppose $a, b, c, d$ are integers greater than $2019$ which are four consecutive quarters of an arithmetic row with $a <b <c <d$ . If $a$ and $d$ are squares of two consecutive natural numbers, then the smallest value of $c-b$ is ...

p7. Given a triangle $ABC$ , with $AB = 6$ , $AC = 8$ and $BC = 10$ . The points $D$ and $E$ lies on the line segment $BC$ . with $BD = 2$ and $CE = 4$ . The measure of the angle $\angle DAE$ is ...

p8. Sequqnce of real numbers $a_1,a_2,a_3,...$ meet $\frac{na_1+(n-1)a_2+...+2a_{n-1}+a_n}{n^2}=1$ for each natural number $n$ . The value of $a_1a_2a_3...a_{2019}$ is ....

p9. The number of ways to select four numbers from $\{1,2,3, ..., 15\}$ provided that the difference of any two numbers at least $3$ is ...

p10. Pairs of natural numbers $(m , n)$ which satisfies $m^2n+mn^2 +m^2+2mn = 2018m + 2019n + 2019$ are as many as ...

p11. Given a triangle $ABC$ with $\angle ABC =135^o$ and $BC> AB$ . Point $D$ lies on the side $BC$ so that $AB=CD$ . Suppose $F$ is a point on the side extension $AB$ so that $DF$ is perpendicular to $AB$ . The point $E$ lies on the ray $DF$ such that $DE> DF$ and $\angle ACE = 45^o$ . The large angle $\angle AEC$ is ...

p12. The set of $S$ consists of $n$ integers with the following properties: For every three different members of $S$ there are two of them whose sum is a member of $S$ . The largest value of $n$ is ....

p13. The minimum value of $\frac{a^2+2b^2+\sqrt2}{\sqrt{ab}}$ with $a, b$ positive reals is ....

p14. The polynomial P satisfies the equation $P (x^2) = x^{2019} (x+ 1) P (x)$ with $P (1/2)= -1$ is ....

p15. Look at a chessboard measuring $19 \times 19$ square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of $k$ coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of $k$ so that the game never ends for any initial square selection is ....

21 replies

parmenides51

Nov 11, 2021

chinawgp

7 minutes ago

Geometry

German_bread 0

14 minutes ago

A semicircle k with radius r is constructed over the line segment ST. Let D be a point on the line segment ST that is different from S and T. The two squares ABCD and DEF G lie in the half-plane of the semicircle such that points B and F lie on the semicircle k and points S, C, D, E, and T lie on a straight line in that order. (Points A and/or G can also lie outside the semicircle if necessary.)
Investigate whether the sum of the areas of the squares ABCD and DEFG depends on the position of point D on the line segment ST.

German math olympiad, class 9, 2022

0 replies

German_bread

14 minutes ago

0 replies

Why is the old one deleted?

EeEeRUT 11

N an hour ago by Mathgloggers

Source: EGMO 2025 P1

For a positive integer $N$ , let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$ . Find all $N \geqslant 3$ such that $\gcd( N, c_i + c_{i+1}) \neq 1$ for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$ . Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$ .

Proposed by Paulius Aleknavičius, Lithuania

11 replies

EeEeRUT

Apr 16, 2025

Mathgloggers

an hour ago

Congruence related perimeter

egxa 2

N 2 hours ago by LoloChen

Source: All Russian 2025 9.8 and 10.8

On the sides of triangle $ABC$ , points $D_1, D_2, E_1, E_2, F_1, F_2$ are chosen such that when going around the triangle, the points occur in the order $A, F_1, F_2, B, D_1, D_2, C, E_1, E_2$ . It is given that
$AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2.$ Prove that the perimeters of the triangles formed by the triplets $AD_1, BE_1, CF_1$ and $AD_2, BE_2, CF_2$ are equal.

2 replies

egxa

Yesterday at 5:08 PM

LoloChen

2 hours ago

number theory

Levieee 7

N 2 hours ago by g0USinsane777

Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers $a$ and $b$ , there cannot exist infinitely many $n \in \mathbb{Z}$ such that
$\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.$
$\textbf{Solution}$

Let
$x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.$
Then,
$x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.$ So:
$x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.$
Therefore,
$\sqrt{(n + a)(n + b)} \in \mathbb{N}.$
Let
$(n + a)(n + b) = k^2.$ Assume $n + a \neq n + b$ . Then we have:
$n + a \mid k \quad \text{and} \quad k \mid n + b,$ or it could also be that $k \mid n + a \quad \text{and} \quad n + b \mid k$ .

Without loss of generality, we take the first case:
$(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.$
Thus,
$k_1 k_2 = \frac{n + b}{n + a}.$
Since $k_1 k_2 \in \mathbb{N}$ , we have:
$k_1 k_2 = 1 + \frac{b - a}{n + a}.$
For infinitely many $n$ , $\frac{b - a}{n + a}$ must be an integer, which is not possible.

Therefore, there cannot be infinitely many such $n$ .

7 replies

Levieee

Yesterday at 7:46 PM

g0USinsane777

2 hours ago

inequalities proplem

Cobedangiu 4

N 2 hours ago by Mathzeus1024

$x,y\in R^+$ and $x+y-2\sqrt{x}-\sqrt{y}=0$ . Find min A (and prove):
$A=\sqrt{\dfrac{5}{x+1}}+\dfrac{16}{5x^2y}$

4 replies

Cobedangiu

Yesterday at 11:01 AM

Mathzeus1024

2 hours ago

3 var inquality

sqing 0

2 hours ago

Source: Own

Let $a,b,c$ be reals such that $a+b+c=0$ and $abc\geq \frac{1}{\sqrt{2}} .$ Prove that
$a^2+b^2+c^2\geq 3$ Let $a,b,c$ be reals such that $a+2b+c=0$ and $abc\geq \frac{1}{\sqrt{2}} .$ Prove that
$a^2+b^2+c^2\geq \frac{3}{ \sqrt[3]{2}}$ $a^2+2b^2+c^2\geq 2\sqrt[3]{4}$

0 replies

sqing

2 hours ago

0 replies

Combinatorics

TUAN2k8 0

2 hours ago

A sequence of integers $a_1,a_2,...,a_k$ is call $k-balanced$ if it satisfies the following properties:
$i) a_i \neq a_j$ and $a_i+a_j \neq 0$ for all indices $i \neq j$ .
$ii) \sum_{i=1}^{k} a_i=0$ .
Find the smallest integer $k$ for which: Every $k-balanced$ sequence, there always exist two terms whose diffence is not less than $n$ . (where $n$ is given positive integer)

0 replies

TUAN2k8

2 hours ago

0 replies

pqr/uvw convert

Nguyenhuyen_AG 4

N 2 hours ago by SunnyEvan

Source: https://github.com/nguyenhuyenag/pqr_convert

Hi everyone,
As we know, the pqr/uvw method is a powerful and useful tool for proving inequalities. However, transforming an expression $f(a,b,c)$ into $f(p,q,r)$ or $f(u,v,w)$ can sometimes be quite complex. That's why I’ve written a program to assist with this process.
I hope you’ll find it helpful!

Download: pqr_convert

Screenshot:
IMAGE
IMAGE

4 replies

Nguyenhuyen_AG

Today at 3:39 AM

SunnyEvan

2 hours ago

A nice lemma about incircle and his internal tangent

manlio 0

2 hours ago

Have you a nice proof for this lemma?
Thnak you very much

0 replies

manlio

2 hours ago

0 replies

Nice problem about a trapezoid

manlio 0

2 hours ago

Have you a nice solution for this problem?
Thank you very much

0 replies

manlio

2 hours ago

0 replies

IHC 10 Q25: Eight countries participated in a football tournament

xytan0585 0

2 hours ago

Source: International Hope Cup Mathematics Invitational Regional Competition IHC10

Eight countries sent teams to participate in a football tournament, with the Argentine and Brazilian teams being the strongest, while the remaining six teams are similar strength. The probability of the Argentine and Brazilian teams winning against the other six teams is both $\frac{2}{3}$ . The tournament adopts an elimination system, and the winner advances to the next round. What is the probability that the Argentine team will meet the Brazilian team in the entire tournament?

$A$ . $\frac{1}{4}$

$B$ . $\frac{1}{3}$

$C$ . $\frac{23}{63}$

$D$ . $\frac{217}{567}$

$E$ . $\frac{334}{567}$

0 replies

xytan0585

2 hours ago

0 replies

New geometry problem

titaniumfalcon 4

N Apr 13, 2025 by titaniumfalcon

Post any solutions you have, with explanation or proof if possible, good luck!

4 replies

titaniumfalcon

Apr 3, 2025

titaniumfalcon

Apr 13, 2025

New geometry problem

G H J

Reveal topic tags

geometry

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titaniumfalcon

115 posts

titaniumfalcon

#1 h Apr 3, 2025, 10:40 PM

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Post any solutions you have, with explanation or proof if possible, good luck!

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mathprodigy2011

313 posts

mathprodigy2011

#2 h Apr 4, 2025, 1:51 AM

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I think the answer is 4sqrt2 but I think there's a flaw in my solution(please point it out if possible):

It's pretty obvious that CA is perpendicular to AB. Let CH be length y, AB=CD=x and the radius of the circle be r. Connect C with the tangent from H and let the point be P.

First note that CAB is a right triangle so r^2+x^2=2y^2 (1)(because BC = sqrt(2)*CH)

Then note that CPH is a right triangle so r^2+4^2=y^2 ===> r^2+16=y^2 (2)

Now let the point where the tangent from D to the circle be point Q. DCQ is a right triangle. The hypotenuse is x and one leg with side length r. So we need to find x^2-r^2. We can do this from out systems. Plug in our equation for y^2 into equation (1). r^2+x^2 = 2r^2+32 or x^2-r^2=32.

Therefore, the tangent has length sqrt(32) or 4sqrt2

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fruitmonster97

2477 posts

fruitmonster97

#3 h Apr 4, 2025, 12:06 PM • 1 Y

Y by titaniumfalcon

w problem

@above is correct.
solution

The power of $H$ is $CH^2-CA^2,$ so $CH^2-CA^2=16,$ and thus $BC^2-2AC^2=32.$ Thus, $AB^2-AC^2=32.$ The power of $D$ is $CD^2-CA^2=AB^2-AC^2=32,$ so the tangent length is $\sqrt{32}=\boxed{4\sqrt2}.$

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mathprodigy2011

313 posts

mathprodigy2011

#4 h Apr 4, 2025, 11:16 PM

Y by

fruitmonster97 wrote:

w problem

@above is correct.
solution

The power of $H$ is $CH^2-CA^2,$ so $CH^2-CA^2=16,$ and thus $BC^2-2AC^2=32.$ Thus, $AB^2-AC^2=32.$ The power of $D$ is $CD^2-CA^2=AB^2-AC^2=32,$ so the tangent length is $\sqrt{32}=\boxed{4\sqrt2}.$

bro this is so smart

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titaniumfalcon

115 posts

titaniumfalcon

#5 h Apr 13, 2025, 4:40 PM

Y by

fruitmonster97 wrote:

w problem

@above is correct.
solution

The power of $H$ is $CH^2-CA^2,$ so $CH^2-CA^2=16,$ and thus $BC^2-2AC^2=32.$ Thus, $AB^2-AC^2=32.$ The power of $D$ is $CD^2-CA^2=AB^2-AC^2=32,$ so the tangent length is $\sqrt{32}=\boxed{4\sqrt2}.$

Thanks! This was my intended solution.

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