OMPR 3d Round 2022, Middle School

Problem 1

A subset of $10$ elements of the set $\{1,2,3,\dots ,90\}$ is called Boricua if by placing its elements in increasing order we have the following property: the difference between the second and the first is $1$, the

third and second is $2$, the fourth and third is $3$, the fifth with the fourth is $4$, etc. How many Boricua subsets are there?

Solution

Let $B$ be a Boricua set and let $x$ be the smallest element of $B$. Then the elements of $B$ are fixed with the largest element being $x+1+2+\dots +8+9=x+45$. Therefore, $x+45\le 90$, or equivalently $x\le 45$. From here the $x$ can be any integer number from the interval $[1,45]$ and the number of Boricua sets is $\boxed{{\displaystyle 45}}$

Problem 2

How many $3$-digit positive numbers are there which can be represented as a sum of exactly $9$ different powers of $2$?

Solution

Let $x$ be any such $3$-digit number and $M$ be the largest power of $2$ in its representation. Notice that for $M\ge 10$ we have

$$x>2^M\ge 2^{10}>1000$$

which contradicts the fact that $x$ is a $3$-digit number. From here $M\le 9$.

Let us consider the list of $10$ numbers

$$2^0,2^1,2^2,\dots ,2^7,2^8,2^9$$

The representation can be obtained by eliminating one of the powers from the list. It is not hard to see that we obtain $3$-digits only by the elimination of $2^9$, $2^8$, $2^7$, $2^6$, and $2^5$. Therefore, the number of such $3$-digit numbers is $\boxed{{\displaystyle 5}}$

Problem 3

Circles ${\omega }_2$, ${\omega }_3$, ${\omega }_4$ are tangent to the circle ${\omega }_1$ externally. Circles ${\omega }_2$ are ${\omega }_3$ are tangent with each other but are not tangent with the circle ${\omega }_4$. Using three different colors, in how many different ways can the following circles be colored so that NO two of the same color touch each other?

Solution

Let ${\omega }_1$ be colored with the color $A$, which can be chosen in $3$ ways. Then ${\omega }_4$ can be colored with any of the $2$ remaining colors: $B$ or $C$, which can be chosen in $2$ ways. If ${\omega }_2$ is colored in $B$, then ${\omega }_3$ should be colored in $C$, and vice versa, which can be chosen in $2$ ways. Therefore, the total number of colorings is $3\cdot 2\cdot 2=\boxed{{\displaystyle 12}}$

Problem 4

Consider the integers $x$ and $y$:

$$x={10}^{2022}+{10}^{2021}+{10}^{2020}+\dots +{10}^2+10+1$$

$$y={10}^{2023}+{10}^{2022}+{10}^{2021}+\dots +{10}^2+10+1$$

What is the sum of the digits of $x+y$?

Solution

Notice that $x$ and $y$ are of the form $11\dots 1$, where digit $1$ is used $2023$ and $2024$ times respectively. The number $x+y$ is of the form $122\dots 2$, where there are $2023$ where digit $2$ is used $2023$ times. Therefore, the sum of digits of $x+y$ is equal to $1+2\cdot 2023=\boxed{{\displaystyle 4047}}$

Problem 5

Let $ABCD$ be a quadrilateral, such that $BC=CD$, $\mathrm{\angle }ABC={90}^{\circ }$, $\mathrm{\angle }ACD={90}^{\circ }$, $\mathrm{\angle }AOD={100}^{\circ }$, where $O$ is the point of intersection of its diagonals. Calculate the measure of the angle $\mathrm{\angle }BAC$.

Solution

Let $\mathrm{\angle }BAC=2x$. Then $\mathrm{\angle }BCA={90}^{\circ }–2x$ and $\mathrm{\angle }CDB=\mathrm{\angle }CBD=x$. From the triangle $BOC$:

$${100}^{\circ }+x+{90}^{\circ }–2x={180}^{\circ }$$

From here $x={10}^{\circ }$ and $\mathrm{\angle }BAC=\boxed{{\displaystyle {20}^{\circ }}}$

Problem 6

Let $S$ be the set of points $(i,j)$, where $i,j\in \{1,2,3,4\}$. By connecting four points from $S$, how many squares of different sizes can be formed?

Solution

It is not hard to see how to obtain the squares of sides $1$, $2$, and $3$.

It is also possible to obtain the squares of the side $\sqrt{2}$, for example, by connecting the vertices $(2,1)$, $(3,2)$, $(2,3)$, and $(2,1)$, and the squares of the side $\sqrt{5}$, for example, by connecting the vertices $(2,1)$, $(4,2)$, $(3,4)$, $(1,3)$. Therefore, the number of squares of different sizes is $\boxed{{\displaystyle 5}}$

Problem 7

A necklace is made up of $2022$ stones, $3$ of which are black and the rest are white. If numbered as shown in the figure, the black stones are at positions $1$, $674$, and $1348$. A move consists of exchanging a black stone with any white stone next to it. What is the minimum number of moves needed for the three black stones to be together?

Solution

Since initially the numbers of consecutive white stones are $672$, $673$, and $674$, then we should move the stones on the positions $1$ and $1348$ towards the stone on the position $674$. The total number of the moves, therefore, is equal to $672+673=\boxed{{\displaystyle 1345}}$

Problem 8

Find all two-digit numbers $\overline{ab}$ such that

$$\frac{\overline{ab}}{4}=\frac{\overline{ba}}{7}$$

Solution

Using the decimal representation of the numbers we have

$$\frac{10a+b}{4}=\frac{10b+a}{7}$$

which, after cross-multiplication, implies that $2a=b$. Since $0<b<10$, then $a$ can only take values $1$, $2$, $3$ and $4$, which provides the numbers $\boxed{{\displaystyle 12,24,36,48}}$

 

Problem 9

$M$ is the midpoint of the side $OB$ of the right triangle $AOB$, such that $OA=OB=12$ and $\mathrm{\angle }AOB={90}^{\circ }$. Let $C$ a point, such that $CM\perp OB$, $CM=OM$ and $C$ and $A$ lie in different semiplanes with respect to $OB$. Let $q_1$ and $q_2$ be the quarter circles with centers $O$ and $M$ and the arcs $AB$ and $BC$ respectively. Find the area of the region formed by the segment $AC$ and the arcs $AB$ and $BC$.

Solution

Notice that $q_1$ is of radius $12$ and its area is $36\pi$, and $q_2$ is of radius $6$ and its area is $9\pi$.

Let $X$ be the intersection of $AC$ and $OB$ and let $a=XM$. From the similarity of the triangles $CMX$ and $AOX$ we have

$$\frac{CM}{OA}=\frac{a}{6-a}$$

From here $a=2$, the area of the triangle $CMX$ is $6$ and the area of the triangle $AOX$ is $24$. Therefore, the area of the needed region is equal to $36\pi +9\pi +6–24=\boxed{{\displaystyle 45\pi -18}}$

Problem 10

In the right triangle $ABC$ ($\mathrm{\angle }C={90}^{\circ }$), the segment $AD$ is perpendicular to the segment $BC$, $AC=6$ and $BD=5$. What is the area of triangle $ABC$?

Solution

Let $AD=x$. From the similarity of the triangles $ADB$ and $ACD$:

$$\frac{x}{5}=\frac{CD}{x}$$

and $CD=\frac{x^2}{5}$.

From the Pythagorean Theorem to the triangle $ACD$:

$$x^2+\frac{x^4}{25}=6^2$$

which implies that $x=2\sqrt{5}$.

From here $CD=4$, $BC=9$, and the area of the triangle $ABC$ is equal to

$$\frac{BC\cdot AD}{2}=\frac{9\cdot 2\sqrt{5}}{2}=\boxed{{\displaystyle 9\sqrt{5}}}$$