OBMEP 2023 – 1st phase – Level 3: 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 — Overlap

And it was glued last. A is below it and above B; B covers C and D. Therefore the first two are C and D, the third is B, the fourth A and the fifth E. Alternative B.

Question 2 — Path and direct distance

Each deviation in the figure covers three sides of a square, while its projection on AB measures only one side. If these sides measure $a,b,\ldots,g$, the distance is $3(a+b+\cdots+g)=6$ km. The direct distance AB is the sum without the factor 3: 2 km — B.

Question 3 — Ball distribution

Two consecutive boxes receive $2+3=5$ balls. The first 404 pairs keep 2020 balls in 808 boxes. Box 809 receives 2021 and 2022; the next receives 2023, 2024 and 2025. The ball 2023 is in the box 810 — E.

Question 4 — Difference in products

Use $u=20235$ and the product of the sum and difference:

Alternative C. Identity avoids two extensive multiplications.

Question 5 — Travel distance

In four hours at 60 km/h, 240 km are covered. If the two complete sections measure $x$ and $D-x$, the part traveled between the indicated times is $3x/4+3(D-x)/4=3D/4$. Therefore $3D/4=240$, and $D=320$ km — D.

Question 6 — Equation and inverse

As $x^2-3x+1=0$, we have $x\ne0$. Dividing the equation by $x$, we obtain $x+1/x=3$. Square:

Alternative A. The result is valid for both roots.

Question 7 — Area of ​​a square

The yellow region is divided into triangles with areas $3l^2$ and $2l^2$, calculated by the bases and heights in the mesh. As the sum is 30, $5l^2=30$. The area of ​​each square is $6\text{ cm}^2$ — E.

Question 8 — Product with radicals

Group the eight factors by pairs that differ only in the sign of 1. Putting $s=\sqrt2+\sqrt3$ and $t=\sqrt2-\sqrt3$, the complete product is $(s^2-1)^2(t^2-1)^2$. Like $s^2-1=4+2\sqrt6$ and $t^2-1=4-2\sqrt6$,

Alternative A.

Question 9 — Identification of ages

Only the 8 year old child lies, and the only sentence that would be false for him is the denial of 8 age. The other sentences are true: the child who denies 9 has 7; the one that denies 7 has 9. The requested identification is in B.

Question 10 — Paths

The two highlighted extreme points admit one path each. Each of the two internal points allows six arrival and six departure routes. The four cases do not overlap, because the displacement only allows going up or forward to the right. Total: $1+36+36+1=74$ — E.

Question 11 — Advantages in races

Comparisons give $v_B/v_A=80/100$ and $v_C/v_B=380/400$. Multiplying,

While André travels 800 m, Carlos travels $800\cdot19/25=608$ m. The required advantage is $800-608=192$ m — E. Adding the two original advantages would ignore the different race distances.

Question 12 — Tangent circles

Call the outer and inner radii $R$ and $r$. The center of the small circle is at angle bisector angle 60°. The triangle formed with the point of tangency has an angle of 30°, so the distance from this center to the vertex is $2r$. Due to the internal tangency and alignment of the centers, this same distance is $2R-r$. So $2R-r=2r$, or $R=3r/2$.

The ratio between the blue area and the pink area is

Alternative B.

Question 13 — Squares and Pythagoras

If the white squares have sides $L$ and the indicated offset measures $x$, the blue sides are $L+x$ and $L-x$. The sum of the blue areas is $2L^2+2x^2$. The highlighted right triangle has hypotenuse 2, therefore $L^2+x^2=4$. Therefore the sum is 8 — D.

Question 14 — Pairs with prohibited sum

The 66 numbers from 10 to 76 different from 43 form 33 pairs whose sum is 86. We can only use one per pair. The number 43 and 23 numbers 77 to 99 do not create another prohibited pair. The maximum is $33+1+23=57$, reached by choosing from 43 to 99. Alternative C.

Question 15 — Reorganization of areas

The segments of 10 cm and the angles of 60° allow parallelograms to be formed where the pieces meet. Its diagonals cut each other in half; thus, each triangle that leaves the central quadrilateral has a congruent triangle that fills an indentation. Transfer these pairs without changing the area. The resulting contour is a square with side 20 cm: the internal angles are right by the sum of the angles of 30° and 60° indicated in the construction. The area is $20^2=400\text{ cm}^2$ — D.

Question 16 — Sum that becomes a product

The sum of 1 to 17 is $17\cdot18/2=153$. Removing $a$ and $b$, the condition is $153-a-b=ab$. Add 1 and factor:

Both factors are between 2 and 18. Among the pairs of dividers of $154=2\cdot7\cdot11$, only $11\cdot14$ responds. So $\{a,b\}=\{10,13\}$ and the requested product is 130 — D.

Question 17 — Numbers on faces

If the product on opposite faces is constant, the numbers opposite 1, 3 and 6 are $6t,2t,t$, for some positive integer $t$. Product 98 in a view cannot use faces 3 and 6, as 98 is not a multiple of 3. You also cannot use $6t$. Therefore it uses $1,2t,t$, and $2t^2=98$ gives $t=7$. The six numbers are $1,3,6,42,14,7$; the largest is 42 — E.

Question 18 — Blocks of people

Conditioned to stay together, the five girls can occupy a block starting at any position from 1 to 6. In each position there are $5!\cdot5!$ ways to order girls and boys, so the six cases have the same weight. The boys only stay together if the girls’ block is at one end. The probability is $2/6=1/3$ — C.

Question 19 — Strip folds

Denote the length by $a$ and the width by $b$. In square overlay, the side measures $b$ and its diagonal measures $b\sqrt2$. Due to the symmetry of the fold, half the length is the sum of this diagonal with the section of length $2b$ obtained in the adjacent isosceles right triangle. Therefore,

Alternative B. Distinguish the side of the overlapping square from its diagonal: they are the ones that give rise to the factors 1 and $\sqrt2$.

Question 20 — Tournament with withdrawals

If $N$ players remain, they play $N(N-1)/2$ matches against each other. Let $k$ be the number of matches between the three dropouts, with $0\le k\le3$. The two appearances of each dropout add up to 6, but a match between two of them was counted twice. Therefore, the additional total is $6-k$ and

We need a triangular number between 44 and 47; the only one is 45, with $N=10$ and $k=1$. Adding up the three dropouts, there were 13 players — D.

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