OBMEP 2023 – 1st phase – Level 2: complete resolution

Official material: complete proof · OBMEP solutions · collection of the year.

The resolutions below follow the numbering of the official notebook. Open the test to follow the statements, alternatives and diagrams; The explanations here are copyright. Attempt each question before checking the answer.

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Question 1 — Pins

The bottom left and bottom right sheets must keep their respective pins: no other holds them. It is also necessary to keep at least one pin from the top left sheet and one from the top right. So that leaves at least four of the ten. Choose, on these two upper sheets, pins that also secure the overlapping sheets in the middle. This choice supports the six sheets using four pins. We can remove $10-4=6$ — E.

Question 2 — Overlapping cards

And he is above everyone; A comes just below it and above B. B, in turn, covers the two initial cards, C and D. The partial order is $(C,D)$, B, A, E, with the possibility of exchanging the first two. The third is B — alternative B.

Question 3 — Pizzas

With $k$ pizzas cut into eight slices, the other $5-k$ have six. The total is $30+2k$, which takes on 30, 32, 34, 36, 38, or 40. Only 36 is a multiple of 9, giving four slices per friend — A.

Question 4 — Boards

Each board uses $7\cdot17=119$ pieces. Like $2023=7\cdot17^2$, dividing by 119 results in 17 boards — E. The area of ​​each piece is $1\text{ cm}^2$, so the count coincides with the area covered.

Question 5 — Reading on alternate days

The 21 pages correspond to three sessions: Friday, Sunday and Tuesday. There are seven pages per session. The 105 page book requires 15 sessions; between the first and last there are 14 intervals of two days, or 28 days. Four weeks after a Friday it is Friday — C again.

Question 6 — Blue cubes

Count the cubes that have at least one white face: there are nine, and none of them can be completely blue. Of the 14 cubes in the stack, the other five can have hidden blue faces without contradicting the design. Answer: five — E.

Question 7 — Ratio between areas of squares

Draw the diagonal AC of the larger square, the diagonals of the smaller one and the perpendiculars to the bases of the side triangles. Using the equal segments indicated in the drawing, one half of the large square breaks down into nine triangles congruent to XBY. The entire square has 18 triangular units, and the smallest has four. Therefore,

Alternative A. Comparing just the sides drawn in different directions does not provide the ratio between areas.

Question 8 — Ferris wheel

Diametrically opposite points have heights whose average is the height of the center. Therefore, the sum of the heights remains constant, even when the wheel rotates. Initially the sum is $3+5=8$ m. When Mary is 7 m, Paul is $8-7=1$ m — B.

Question 9 — Area on the mesh

If each square has a side $l$, divide the yellow region into the two triangles indicated by the vertices of the drawing. Their bases and heights, measured on the mesh, give areas $(2l)(3l)/2=3l^2$ and $(2l)(2l)/2=2l^2$. Then $5l^2=30$, whence $l^2=6\text{ cm}^2$ — E.

Question 10 — Products and inverses

The three products of the inverses of the unknowns are 2, 3 and 6. To obtain the square of an inverse, multiply the two products that contain it and divide by the third. The three squares are, in some order,

The requested sum is worth $1+4+9=14$ — C. This procedure does not require finding the signs of the unknowns.

Question 11 — Divisibility of a three-digit number

Write $\overline{ABC}=10\overline{AB}+C$. When dividing by $\overline{AB}$, the remaining part is $C/\overline{AB}$. Like $0\le C\le9<\overline{AB}$, it can only be integer when $C=0$. For A there are nine non-zero choices; for B, eight, excluding zero and A. Total: $9\cdot8=72$ — C.

Question 12 — Sum and product

Like $2048=2^{11}$, the four positive integers are powers of 2. The sum 67 is odd, so there are one or three terms equal to 1. With three ones, 64 remains, but the product would be 64. With just one 1, the other three powers add up to 66: they must be 32, 32 and 2, because, if they were all at maximum 16, the sum would not reach 66; a term 64 would leave two even positives adding 2, impossible. The product is $32\cdot32\cdot2\cdot1=2048$. The smallest sum of three is $32+2+1=35$ — B.

Question 13 — Ages and lies

The 8 year old needs to make a false statement. Of the three sentences, only “it’s not 8” is false for her. The others tell the truth: whoever denies having 9 has 7, and whoever denies having 7 has 9. The oldest and youngest are those indicated in B.

Question 14 — Paths in the mesh

At the two highlighted extreme points, there is one path per point. For each internal point, there are six ways to get there and six ways to go to B. Since the paths up and to the right cannot pass through two of these points, there is no double counting. The total is $1+1+6^2+6^2=74$ — E.

Question 15 — Card that forms a box

Denote by $h$ the height of the box and by $b,c$ the sides of the base. We have $hb=28$, $hc=24$ and $hbc=168$. So $c=168/28=6$, $h=24/6=4$ and $b=28/4=7$. Before cuts, the card measures $(b+2h)\times(c+2h)$, that is, $15\times14$. Its area is $210\text{ cm}^2$ — D.

Question 16 — Rhombuses

Each rhombus is the union of two neighboring small triangles. There are three possible orientations. In one orientation, count in ranges: $4+3+2+1=10$. A rotation of the large triangle takes this orientation to each of the other two, without changing the count. Then there is $3\cdot10=30$ — B.

Question 17 — Different ages

If the greatest age were at most 15, the greatest sum of ten distinct integer ages would be $6+7+\cdots+15=105$, insufficient. Therefore Mariana has at least 16. This limit is possible with $6,7,8,9,10,12,13,14,15,16$, whose sum is 110. Therefore, the minimum is 16 years — D.

Question 18 — Names and colors

The girl in blue is not Alice, Cláudia or Daniela; So it’s Beatriz. Cláudia doesn’t wear black or red according to the second information, and she doesn’t wear blue either: she wears white. In the circle, Beatriz stands between Cláudia and Daniela. The girl in black needs to be between the girl in red and Cláudia, so it cannot be Daniela, who is Beatriz’s neighbor. We conclude: Alice in black, Beatriz in blue, Cláudia in white and Daniela in red — E.

Question 19 — Avoiding sum 86

From 10 to 76, form 33 pairs $(10,76),(11,75),\ldots,(42,44)$. We can choose a maximum of one of each pair. 43 can enter, as the same number is not repeated; all 23 numbers from 77 to 99 can also. The limit is $33+1+23=57$, reached by choosing all numbers from 43 to 99. Response: 57 — C.

Question 20 — Area of ​​the star

When decomposing the drawing, use the rhombus ABCD: the diagonal BD divides it into two triangles of the same area. Furthermore, triangles PBC and QCD are congruent along the sides of the regular pentagon. The blue and white parts each have three triangles of the first type and one of the second. The areas are equal; the blue fraction is $1/2$ — C.

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