Maths Olympiad Prep

Track / Stage 6 / 276 of 400 #1276 of 1964

Problem 1276

National olympiad, first round
Combinatorics Difficulty 6.5 Find the answer

The following isosceles trapezoid consists of equal equilateral triangles with side length 11. The side A1EA_1E has length 33 while the larger base A1AnA_1A_n has length n1n-1. Starting from the point A1A_1 we move along the segments which are oriented to the right and up(obliquely right or left). Calculate (in terms of nn or not) the number of all possible paths we can follow, in order to arrive at points B,Γ,Δ,EB,\Gamma,\Delta, E, if nn is an integer greater than 33.

[Need image]

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

1. Define the structure of the isosceles trapezoid:
- The trapezoid consists of equal equilateral triangles with side length 1.
- The side A1E A_1E has length 3.
- The larger base A1An A_1A_n has length n1 n-1 .

2. Identify the layers:
- The trapezoid can be divided into layers, where each layer is a horizontal section between two lines.
- The bottom layer has 2n2 2n-2 oblique segments.
- The middle layer has 2n4 2n-4 oblique segments.
- The top layer has 2n6 2n-6 oblique segments.

3. **Calculate the number of paths to point E E :**
- To go from A1 A_1 to E E , we need to choose one oblique segment from each layer.
- The number of ways to choose these segments is:
(2n2)(2n4)(2n6) (2n-2)(2n-4)(2n-6)
- Simplify the expression:
(2n2)(2n4)(2n6)=8n348n2+88n48 (2n-2)(2n-4)(2n-6) = 8n^3 - 48n^2 + 88n - 48

4. **Calculate the number of paths to point Δ \Delta :**
- The number of paths to Δ \Delta is the same as to E E :
8n348n2+88n48 8n^3 - 48n^2 + 88n - 48

5. **Calculate the number of paths to point Γ \Gamma :**
- To go from A1 A_1 to Γ \Gamma , we only need to choose oblique segments from the bottom and middle layers.
- The number of ways to choose these segments is:
(2n2)(2n4) (2n-2)(2n-4)
- Simplify the expression:
(2n2)(2n4)=4n212n+8 (2n-2)(2n-4) = 4n^2 - 12n + 8

6. **Calculate the number of paths to point B B :**
- To go from A1 A_1 to B B , we only need to choose oblique segments from the bottom layer.
- The number of ways to choose these segments is:
2n2 2n-2

7. Sum the total number of paths:
- The total number of paths is the sum of the paths to E E , Δ \Delta , Γ \Gamma , and B B :
2(8n348n2+88n48)+(4n212n+8)+(2n2) 2(8n^3 - 48n^2 + 88n - 48) + (4n^2 - 12n + 8) + (2n - 2)
- Simplify the expression:
2(8n348n2+88n48)+4n212n+8+2n2=16n396n2+176n96+4n212n+8+2n2 2(8n^3 - 48n^2 + 88n - 48) + 4n^2 - 12n + 8 + 2n - 2 = 16n^3 - 96n^2 + 176n - 96 + 4n^2 - 12n + 8 + 2n - 2
=16n392n2+166n90 = 16n^3 - 92n^2 + 166n - 90

The final answer is 16n392n2+166n90\boxed{16n^3 - 92n^2 + 166n - 90}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.