Found problems: 85335
2005 Bosnia and Herzegovina Team Selection Test, 1
Let $H$ be an orthocenter of an acute triangle $ABC$. Prove that midpoints of $AB$ and $CH$ and intersection point of angle bisectors of $\angle CAH$ and $\angle CBH$ lie on the same line.
2013 Bosnia And Herzegovina - Regional Olympiad, 3
Find all integers $a$ such that $\sqrt{\frac{9a+4}{a-6}}$ is rational number
2018 China Team Selection Test, 6
Find all pairs of positive integers $(x, y)$ such that $(xy+1)(xy+x+2)$ be a perfect square .
2022 Turkey EGMO TST, 5
We are given three points $A,B,C$ on a semicircle. The tangent lines at $A$ and $B$ to the semicircle meet the extension of the diameter at points $M,N$ respectively. The line passing through $A$ that is perpendicular to the diameter meets $NC$ at $R$, and the line passing through $B$ that is perpendicular to the diameter meets $MC$ at $S$. If the line $RS$ meets the extension of the diameter at $Z$, prove that $ZC$ is tangent to the semicircle.
2019 Purple Comet Problems, 4
Of the students attending a school athletic event, $80\%$ of the boys were dressed in the school colors, $60\%$ of the girls were dressed in the school colors, and $45\% $ of the students were girls. Find the percentage of students attending the event who were wearing the school colors.
2022 Stanford Mathematics Tournament, 1
George is drawing a Christmas tree; he starts with an isosceles triangle $AB_0C_0$ with $AB_0=AC_0=41$ and $B_0C_0=18$. Then, he draws points $B_i$ and $C_i$ on sides $AB_0$ and $AC_0$, respectively, such that $B_iB_{i+1}=1$ and $C_iC_{i+1}=1$ ($B_{41}=C_{41}=A$). Finally, he uses a green crayon to color in triangles $B_iC_iC_{i+1}$ for $i$ from $0$ to $40$. What is the total area that he colors in?
2007 All-Russian Olympiad Regional Round, 11.5
Find all positive integers $ n$ for which there exist integers $ a,b,c$ such that $ a\plus{}b\plus{}c\equal{}0$ and the number $ a^{n}\plus{}b^{n}\plus{}c^{n}$ is prime.
1987 Romania Team Selection Test, 4
Let $ P(X) \equal{} a_{n}X^{n} \plus{} a_{n \minus{} 1}X^{n \minus{} 1} \plus{} \ldots \plus{} a_{1}X \plus{} a_{0}$ be a real polynomial of degree $ n$. Suppose $ n$ is an even number and:
a) $ a_{0} > 0$, $ a_{n} > 0$;
b) $ a_{1}^{2} \plus{} a_{2}^{2} \plus{} \ldots \plus{} a_{n \minus{} 1}^{2}\leq\frac {4\min(a_{0}^{2} , a_{n}^{2})}{n \minus{} 1}$.
Prove that $ P(x)\geq 0$ for all real values $ x$.
[i]Laurentiu Panaitopol[/i]
2010 India Regional Mathematical Olympiad, 2
Let $P_1(x) = ax^2 - bx - c$, $P_2(x) = bx^2 - cx - a$, $P_3(x) = cx^2 - ax - b$ be three quadratic polynomials. Suppose there exists a real number $\alpha$ such that $P_1(\alpha) = P_2(\alpha) = P_3(\alpha)$. Prove that $a = b = c$.
2024-IMOC, A2
Given integer $n \geq 3$ and $x_1$, $x_2$, …, $x_n$ be $n$ real numbers satisfying $|x_1|+|x_2|+…+|x_n|=1$. Find the minimum of
\[|x_1+x_2|+|x_2+x_3|+…+|x_{n-1}+x_n|+|x_n+x_1|.\]
[i]Proposed by snap7822[/i]
2003 Estonia National Olympiad, 2
Find all positive integers $n$ such that $n+ \left[ \frac{n}{6} \right] \ne \left[ \frac{n}{2} \right] + \left[ \frac{2n}{3} \right]$
2008 IMAR Test, 3
Two circles $ \gamma_{1}$ and $ \gamma_{2}$ meet at points $ X$ and $ Y$. Consider the parallel through $ Y$ to the nearest common tangent of the circles. This parallel meets again $ \gamma_{1}$ and $ \gamma_{2}$ at $ A$, and $ B$ respectively. Let $ O$ be the center of the circle tangent to $ \gamma_{1},\gamma_{2}$ and the circle $ AXB$, situated outside $ \gamma_{1}$ and $ \gamma_{2}$ and inside the circle $ AXB.$ Prove that $ XO$ is the bisector line of the angle $ \angle{AXB}.$
[b]Radu Gologan[/b]
2003 All-Russian Olympiad, 2
The diagonals of a cyclic quadrilateral $ABCD$ meet at $O$. Let $S_1, S_2$ be the circumcircles of triangles $ABO$ and $CDO$ respectively, and $O,K$ their intersection points. The lines through $O$ parallel to $AB$ and $CD$ meet $S_1$ and $S_2$ again at $L$ and $M$, respectively. Points $P$ and $Q$ on segments $OL$ and $OM$ respectively are taken such that $OP : PL = MQ : QO$. Prove that $O,K, P,Q$ lie on a circle.
2010 Purple Comet Problems, 10
A baker uses $6\tfrac{2}{3}$ cups of flour when she prepares $\tfrac{5}{3}$ recipes of rolls. She will use $9\tfrac{3}{4}$ cups of flour when she prepares $\tfrac{m}{n}$ recipes of rolls where m and n are relatively prime positive integers. Find $m + n.$
2006 MOP Homework, 1
In how many ways can the set $N ={1, 2, \cdots, n}$ be partitioned in the form $p(N) = A_{1}\cup A_{2}\cup \cdots \cup A_{k},$ where $A_{i}$ consists of elements forming arithmetic progressions, all with the same common positive difference $d_{p}$ and of length at least one? At least two?
[hide="Solution"]
[b]Part 1[/b]
Claim: There are $2^{n}-2$ ways of performing the desired partitioning.
Let $d(k)$ equal the number of ways $N$ can be partitioned as above with common difference $k.$ We are thus trying to evaluate
$\sum_{i=1}^{n-1}d(i)$
[b]Lemma: $d(i) = 2^{n-i}$[/b]
We may divide $N$ into various rows where the first term of each row denotes a residue $\bmod{i}.$ The only exception is the last row, as no row starts with $0$; the last row will start with $i.$ We display the rows as such:
$1, 1+i, 1+2i, 1+3i, \cdots$
$2, 2+i, 2+2i, 2+3i, \cdots$
$\cdots$
$i, 2i, 3i, \cdots$
Since all numbers have one lowest remainder $\bmod{i}$ and we have covered all possible remainders, all elements of $N$ occur exactly once in these rows.
Now, we can take $k$ line segments and partition a given row above; all entries within two segments would belong to the same set. For example, we can have:
$1| 1+i, 1+2i, 1+3i | 1+4i | 1+5i, 1+6i, 1+7i, 1+8i,$
which would result in the various subsets: ${1},{1+i, 1+2i, 1+3i},{1+4i},{1+5i, 1+6i, 1+7i, 1+8i}.$ For any given row with $k$ elements, we can have at most $k-1$ segments. Thus, we can arrange any number of segments where the number lies between $0$ and $k-1$, inclusive, in:
$\binom{k-1}{0}+\binom{k-1}{1}+\cdots+\binom{k-1}{k-1}= 2^{k-1}$
ways. Applying the same principle to the other rows and multiplying, we see that the result always gives us $2^{n-i},$ as desired.
We now proceed to the original proof.
Since $d(i) = 2^{n-i}$ by the above lemma, we have:
$\sum_{i=1}^{n-1}d(i) = \sum_{i=1}^{n-1}2^{n-i}= 2^{n}-2$
Thus, there are $2^{n}-2$ ways of partitioning the set as desired.
[b]Part 2[/b]
Everything is the same as above, except the lemma slightly changes to $d(i) = 2^{n-i}-i.$ Evaluating the earlier sum gives us:
$\sum_{i=1}^{n-1}d(i) = \sum_{i=1}^{n-1}2^{n-i}-i = 2^{n}-\frac{n(n-1)}{2}-2$
[/hide]
I Soros Olympiad 1994-95 (Rus + Ukr), 10.6
Find all functions $f:R\to R$ such that for any real $x, y$ , $$f(x+2^y)=f(2^x)+f(y)$$
2022 Vietnam TST, 1
Given a real number $\alpha$ and consider function $\varphi(x)=x^2e^{\alpha x}$ for $x\in\mathbb R$. Find all function $f:\mathbb R\to\mathbb R$ that satisfy: $$f(\varphi(x)+f(y))=y+\varphi(f(x))$$ forall $x,y\in\mathbb R$
2013 Korea Junior Math Olympiad, 6
Find all functions $f : \mathbb{N} \rightarrow \mathbb{N} $ satisfying
\[ f(mn) = \operatorname{lcm} (m,n) \cdot \gcd( f(m), f(n) ) \]
for all positive integer $m,n$.
2001 All-Russian Olympiad, 1
The integers from $1$ to $999999$ are partitioned into two groups: the first group consists of those integers for which the closest perfect square is odd, whereas the second group consists of those for which the closest perfect square is even. In which group is the sum of the elements greater?
2011 Korea Junior Math Olympiad, 4
For a positive integer $n$, ($n\ge 2$), find the number of sets with $2n + 1$ points $P_0, P_1,..., P_{2n}$ in the coordinate plane satisfying the following as its elements:
- $P_0 = (0, 0),P_{2n}= (n, n)$
- For all $i = 1,2,..., 2n - 1$, line $P_iP_{i+1}$ is parallel to $x$-axis or $y$-axis and its length is $1$.
- Out of $2n$ lines$P_0P_1, P_1P_2,..., P_{2n-1}P_{2n}$, there are exactly $4$ lines that are enclosed in the domain $y \le x$.
2012 Purple Comet Problems, 1
Evaluate $5^4-4^3-3^2-2^1-1^0$.
2016 Estonia Team Selection Test, 6
A circle is divided into arcs of equal size by $n$ points ($n \ge 1$). For any positive integer $x$, let $P_n(x)$ denote the number of possibilities for colouring all those points, using colours from $x$ given colours, so that any rotation of the colouring by $ i \cdot \frac{360^o}{n}$ , where i is a positive integer less than $n$, gives a colouring that differs from the original in at least one point. Prove that the function $P_n(x)$ is a polynomial with respect to $x$.
2020 Harvard-MIT Mathematics Tournament, 8
Let $P(x)$ be the unique polynomial of degree at most $2020$ satisfying $P(k^2)=k$ for $k=0,1,2,\dots,2020$. Compute $P(2021^2)$.
[i]Proposed by Milan Haiman.[/i]
2024 Malaysian IMO Training Camp, 6
Let $\omega_1$, $\omega_2$, $\omega_3$ are three externally tangent circles, with $\omega_1$ and $\omega_2$ tangent at $A$. Choose points $B$ and $C$ on $\omega_1$ so that lines $AB$ and $AC$ are tangent to $\omega_3$. Suppose the line $BC$ intersect $\omega_3$ at two distinct points, and $X$ is the intersection further away to $B$ and $C$ than the other one.
Prove that one of the tangent lines of $\omega_2$ passing through $X$, is also tangent to an excircle of triangle $ABC$.
[i]Proposed by Ivan Chan Kai Chin[/i]
2013 Chile National Olympiad, 5
A conical surface $C$ is cut by a plane $T$ as shown in the figure on the back of this sheet. Show that $C \cap T$ is an ellipse. You can use as an aid the fact that if you consider the two spheres tangent to $C$ and $T$ as shown in the figure, they intersect $T$ in the bulbs.
[asy]
// calculate intersection of line and plane
// p = point on line
// d = direction of line
// q = point in plane
// n = normal to plane
triple lineintersectplan(triple p, triple d, triple q, triple n)
{
return (p + dot(n,q - p)/dot(n,d)*d);
}
// projection of point A onto line BC
triple projectionofpointontoline(triple A, triple B, triple C)
{
return lineintersectplan(B, B - C, A, B - C);
}
// calculate area of space triangle with vertices A, B, and C
real trianglearea(triple A, triple B, triple C)
{
return abs(cross(A - C, B - C)/2);
}
// calculate incentre of space triangle ABC
triple triangleincentre(triple A, triple B, triple C)
{
return (abs(B - C) * A + abs(C - A) * B + abs(A - B) * C)/(abs(B - C) + abs(C - A) + abs(A - B));
}
// calculate inradius of space triangle ABC
real triangleinradius(triple A, triple B, triple C)
{
return 2*trianglearea(A,B,C)/(abs(B - C) + abs(C - A) + abs(A - B));
}
// calculate excentre of space triangle ABC
triple triangleexcentre(triple A, triple B, triple C)
{
return (-abs(B - C) * A + abs(C - A) * B + abs(A - B) * C)/(-abs(B - C) + abs(C - A) + abs(A - B));
}
// calculate exradius of space triangle ABC
real triangleexradius(triple A, triple B, triple C)
{
return 2*trianglearea(A,B,C)/(-abs(B - C) + abs(C - A) + abs(A - B));
}
unitsize(2 cm);
pair project (triple A, real t) {
return((A.x, A.y*Sin(t) + A.z*Cos(t)));
}
real alpha, beta, theta, t;
real coneradius = 1, coneheight = 3;
real a, b, c;
real[] m, r;
triple A, B, V;
triple ellipsecenter, ellipsex, ellipsey;
triple[] F, O, P, R, W;
path[] ellipse, spherering;
theta = 15;
V = (0,0,-coneheight);
m[1] = sqrt(Cos(theta)^2*coneheight^2 - Sin(theta)^2*coneradius^2)/coneradius;
m[2] = -m[1];
alpha = -aTan(Sin(theta)/m[1]);
beta = -aTan(Sin(theta)/m[2]) + 180;
A = (coneradius*Cos(alpha), coneradius*Sin(alpha), 0);
B = (coneradius*Cos(beta), coneradius*Sin(beta), 0);
W[1] = interp(V,(coneradius,0,0),0.6);
W[2] = interp(V,(-coneradius,0,0),0.4);
O[1] = triangleexcentre(V,W[1],W[2]);
O[2] = triangleincentre(V,W[1],W[2]);
r[1] = triangleexradius(V,W[1],W[2]);
r[2] = triangleinradius(V,W[1],W[2]);
F[1] = projectionofpointontoline(O[1],W[1],W[2]);
F[2] = projectionofpointontoline(O[2],W[1],W[2]);
P[1] = O[1] - (0,0,r[1]*coneradius/sqrt(coneradius^2 + coneheight^2));
P[2] = O[2] - (0,0,r[2]*coneradius/sqrt(coneradius^2 + coneheight^2));
spherering[11] = shift(project(P[1],theta))*yscale(Sin(theta))*arc((0,0),r[1]*coneheight/sqrt(coneradius^2 + coneheight^2),alpha,beta);
spherering[12] = shift(project(P[1],theta))*yscale(Sin(theta))*arc((0,0),r[1]*coneheight/sqrt(coneradius^2 + coneheight^2),beta,alpha + 360);
spherering[21] = shift(project(P[2],theta))*yscale(Sin(theta))*arc((0,0),r[2]*coneheight/sqrt(coneradius^2 + coneheight^2),alpha,beta);
spherering[22] = shift(project(P[2],theta))*yscale(Sin(theta))*arc((0,0),r[2]*coneheight/sqrt(coneradius^2 + coneheight^2),beta,alpha + 360);
ellipsecenter = (W[1] + W[2])/2;
a = abs(W[1] - ellipsecenter);
c = abs(F[1] - ellipsecenter);
b = sqrt(a^2 - c^2);
ellipsex = (W[1] - W[2])/abs(W[1] - W[2]);
ellipsey = (0,1,0);
ellipse[1] = project(ellipsecenter + a*ellipsex, theta);
for (t = 0; t <= 180; t = t + 5) {
ellipse[1] = ellipse[1]--project(ellipsecenter + a*Cos(t)*ellipsex + b*Sin(t)*ellipsey, theta);
}
ellipse[2] = project(ellipsecenter - a*ellipsex, theta);
for (t = 180; t <= 360; t = t + 5) {
ellipse[2] = ellipse[2]--project(ellipsecenter + a*Cos(t)*ellipsex + b*Sin(t)*ellipsey, theta);
}
R[1] = ellipsecenter + 1*ellipsex + ellipsey;
R[2] = ellipsecenter - 1.2*ellipsex + ellipsey;
R[3] = ellipsecenter - 1*ellipsex - ellipsey;
R[4] = ellipsecenter + 1.2*ellipsex - ellipsey;
fill(ellipse[1]--ellipse[2]--cycle, gray(0.9));
draw(yscale(Sin(theta))*Circle((0,0),coneradius));
draw(project(V,theta)--project(A,theta));
draw(project(V,theta)--project(B,theta));
draw(Circle(project(O[1],theta),r[1]));
draw(Circle(project(O[2],theta),r[2]));
draw(spherering[11], dashed);
draw(spherering[12]);
draw(spherering[21], dashed);
draw(spherering[22]);
draw(ellipse[1], dashed);
draw(ellipse[2]);
draw(project(R[1],theta)--interp(project(R[1],theta),project(R[2],theta),0.13));
draw(interp(project(R[1],theta),project(R[2],theta),0.13)--interp(project(R[1],theta),project(R[2],theta),0.76), dashed);
draw(interp(project(R[1],theta),project(R[2],theta),0.76)--project(R[2],theta));
draw(project(R[2],theta)--project(R[3],theta)--project(R[4],theta)--project(R[1],theta));
label("$C$", (-1,0.3));
label("$T$", (1.2,-0.8));
dot(project(F[1],theta));
dot(project(F[2],theta));
//dot("$F_1$", project(F[1],theta));
//dot("$F_2$", project(F[2],theta));
//dot("$O_1$", project(O[1],theta));
//dot("$O_2$", project(O[2],theta));
//dot("$P_1$", project(P[1],theta));
//dot("$V$", project(V,theta));
//dot("$W_1$", project(W[1],theta));
//dot("$W_2$", project(W[2],theta));
[/asy]