Maths Olympiad Prep

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Combinatorics Difficulty 6.7 National olympiad Find the answer

[b]p1. [/b]Ed, Fred and George are playing on a see-saw that is slightly off center. When Ed sits on the left side and George, who weighs 100100 pounds, on the right side, they are perfectly balanced. Similarly, if Fred, who weighs 400400 pounds, sits on the left and Ed sits on the right, they are also perfectly balanced. Assuming the see-saw has negligible weight, what is the weight of Ed, in pounds?

[b]p2.[/b] How many digits does the product 2425382^{42}\cdot 5^{38} have?

[b]p3.[/b] Square ABCDABCD has equilateral triangles drawn external to each side, as pictured. If each triangle is folded upwards to meet at a point EE, then a square pyramid can be made. If the center of square ABCDABCD is OO, what is the measure of OEA\angle OEA?
[img]https://cdn.artofproblemsolving.com/attachments/9/a/39c0096ace5b942a9d3be1eafe7aa7481fbb9f.png[/img]

[b]p4.[/b] How many solutions (x,y)(x, y) in the positive integers are there to 3x+7y=13373x + 7y = 1337 ?

[b]p5.[/b] A trapezoid with height 1212 has legs of length 2020 and 1515 and a larger base of length 4242. What are the possible lengths of the other base?

[b]p6.[/b] Let f(x)=6x+7f(x) = 6x + 7 and g(x)=7x+6g(x) = 7x + 6. Find the value of a such that g1(f1(g(f(a))))=1g^{-1}(f^{-1}(g(f(a)))) = 1.

[b]p7.[/b] Billy and Cindy want to meet at their favorite restaurant, and they have made plans to do so sometime between 1:001:00 and 2:002:00 this Sunday. Unfortunately, they didn’t decide on an exact time, so they both decide to arrive at a random time between 1:001:00 and 2:002:00. Silly Billy is impatient, though, and if he has to wait for Cindy, he will leave after 1515 minutes. Cindy, on the other hand, will happily wait for Billy from whenever she arrives until 2:002:00. What is the probability that Billy and Cindy will be able to dine together?

[b]p8.[/b] As pictured, lines are drawn from the vertices of a unit square to an opposite trisection point. If each triangle has legs with ratio 3:13 : 1, what is the area of the shaded region?
[img]https://cdn.artofproblemsolving.com/attachments/e/9/35a6340018edcddfcd7e085f8f6e56686a8e07.png[/img]

[b]p9.[/b] For any positive integer nn, let f1(n)f_1(n) denote the sum of the squares of the digits of nn. For k2k \ge 2, let fk(n)=fk1(f1(n))f_k(n) = f_{k-1}(f_1(n)). Then, f1(5)=25f_1(5) = 25 and f3(5)=f2(25)=85f_3(5) = f_2(25) = 85. Find f2012(15)f_{2012}(15).

[b]p10.[/b] Given that 20120220122012022012 has 8 8 distinct prime factors, find its largest prime factor.

PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

A number or a short expression. Spacing and $ signs are ignored.

Solution

To find the possible lengths of the other base of the trapezoid, we need to consider the two different orientations of the legs. We will use the Pythagorean theorem to determine the horizontal components of the legs.

1. Case 1: Leg of length 15 slopes to the right

- The horizontal component of the leg of length 20:
202122=400144=256=16 \sqrt{20^2 - 12^2} = \sqrt{400 - 144} = \sqrt{256} = 16
- The horizontal component of the leg of length 15:
152122=225144=81=9 \sqrt{15^2 - 12^2} = \sqrt{225 - 144} = \sqrt{81} = 9
- The length of the other base:
42169=17 42 - 16 - 9 = 17

2. Case 2: Leg of length 15 slopes to the left

- The horizontal component of the leg of length 20 remains the same:
202122=16 \sqrt{20^2 - 12^2} = 16
- The horizontal component of the leg of length 15 remains the same:
152122=9 \sqrt{15^2 - 12^2} = 9
- The length of the other base:
4216+9=35 42 - 16 + 9 = 35

Thus, the possible lengths of the other base are 1717 and 3535.

The final answer is 17,35\boxed{17, 35}.

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