#### Problem 1 [***]

There are unique integers \(a_2, a_3, ... a_7 \) such that

$$\frac{5}{7} = \frac{a_2}{2!} + \frac{a_3}{3!} + \frac{a_4}{4!} + \frac{a_5}{5!} + \frac{a_6}{6!} + \frac{a_7}{7!}$$

where \( 0 <= a_i < i\) for i = 2,3...,7

Find \(a_2, a_3 ... a_7 \)

[AHSME]

1. Remove the fractions.

2. Look at this as a remainder problem first what's the remainder when divided by 7 etc.

2. Look at this as a remainder problem first what's the remainder when divided by 7 etc.

#### Problem 2 [*]

Prove: log(1 + 2 + 3) = log (1) + log(2) + log(3)

[@mathematicsprof]

Remember the log rules.

#### Problem 3 [**]

$$(6x + 5)^2(3x +2)(x+1) = 35$$

[@mathematician_3]

1. Substitution is the key \((6x + 5)^2 = 36x^2 + 60x + 25 = 12(3x^2 + 5x + 2) + 1 = 12(3x + 2)(x + 1) + 1\)

#### Problem 4 [*]

The perimeter of the rectangle ABCD is 28 units E is the midpoint of the side AC and the two arc are tangent to each other. What are the dimensions of the rectangle?

[@five_triangles originally]

Find the implicit triangle and then apply the Pythagorean theorem.

#### Problem 5 [**]

This is a pair for comparison:[@matematik_man]

[@eylem_99]

Both depend on forming 30-60-90 triangles. Look especially at the square root 3 side.

#### Problem 6[**]

[UWSIM program]

1. Take logs to get rid of the exponents i.e. \( \log{x} \cdot \log{3} = \log{y} \cdot \log{4} \)

2. Then you can use substitution for log x or log y.

3. Its hard to see the simplification so it might be useful to substitute things in but essentially this reduces to \( \log{x} \cdot (\frac{\log{4}^2 - \log{3}^2}{\log{4}}) = \log{3}^2 - \log{4}^2 \)

4. The difference of squares terms cancel out and you're left with \( \log{x} = -\log{4} \)

2. Then you can use substitution for log x or log y.

3. Its hard to see the simplification so it might be useful to substitute things in but essentially this reduces to \( \log{x} \cdot (\frac{\log{4}^2 - \log{3}^2}{\log{4}}) = \log{3}^2 - \log{4}^2 \)

4. The difference of squares terms cancel out and you're left with \( \log{x} = -\log{4} \)

#### Problem 7 [**]

A,F,B,E,C on circle O. AC=BD=4, CG=3, AC//FE.

Find the area of AEF

[@five_triangles]

1. Note ACG is a 3-4-5 and BEG is similar to it.

2. Since EF is parallel to AC triangle DEG is also similar to the 3-4-5

3. Set x = DG then EG = 5/3X but x + 4 = 5/3(EG) . Solve and DG = x = 9/4.

4. Drop and altitude from A to EF. This divides into a 3-4-5 similar triangle where you have one side AE and a triangle similar to ABC where you also know all the sides.

5. This gives the altitude and base for AEF.

#### Problem 8 [***]

Let p be a 50 digit prime number, when squared its remainder when divided by 120 is

**not**1. What is \(p^2\mod 120\) ?[AOPS]

1. Factor 120

2. Observe the behavior of squares mod each of the factors.

3. If you're prime then you can never be 0 mod any of the factors.

4. Only 2 cases left.

2. Observe the behavior of squares mod each of the factors.

3. If you're prime then you can never be 0 mod any of the factors.

4. Only 2 cases left.

#### Problem 9 [*]

1. Find a polynomial with integer coefficients that has \(3 + \sqrt{5} \) as one root.2. Find a polynomial with integer coefficients that has \(\sqrt{5} + \sqrt{7} \) as one root.

3. Find a polynomial with integer coefficients that has \(3 + \sqrt{5} + \sqrt{7} \) as one root.

[Mike Lawler]

Use conjugates or square the sides enough to remove the radicals. Also don't forget about shifting a function to the right or left.

#### Problem 10 [*]

Find the smallest positive integer N such that \( \frac{N - 13}{5N + 6} \) is a reducible fraction.[AHSME]

Try using the Euclidean Algorithm to find the gcd between the numerator and denominator.

#### Problem 11 [*]

Find the value of \( \sqrt{100 \cdot 101 \cdot 102 \cdot 103 + 1} \) then find a general formula.

[@matematik_man]

There are a variety of methods that work. Start by looking at much smaller examples i.e. n = 1, 2 etc. Curve fit, use a linear system assume the expression is a square etc.

#### Problem 12 [*]

Find sin(2x)

[@eylem]

1. There are multiple trig identity methods to use. Squaring both sides is nice since (cos(x)+sin(x))^ = 1 + sin(2x)
2. However don't miss the geometric interpretation. There are two pythagorean triples here a 3-4-5 and 7-24-25 made out of it.

#### Problem 13 [*]

Choose an integer bigger than all the coefficients, and evaluate. so, for example, p(9)=26260. Tell me x and p(x) and I'll then guess your polynomial (correctly).How does this work?

[Colin Wright]

- Greedy algorithm works.
- The constraint on the value of x makes the poly unique!

#### Problem 14 [*]

1. Add lines to corners
2. Note all triangles with equal areas: A + B = 16 B + C = 20 C + D = 32
3. Then solve for A + D

#### Problem 15 [*]

[GoGeometry]

This is a good example of the power of shape substitution.

- Fill the holes on the top and bottom triangles with the yellow pieces from the side (1/2 of the square)
- Leaving 2 little squares on the left and right with a diagonal of 1/2 side length and an area of 1/8

#### Problem 16 [*]

The side lengths of the squares are 3 consecutive integers. What's the total area?

[@cshearer41]

The key is finding the dimensions of the rectangle (2x1)

With that in hand you can use the Pythagorean Theorem.

#### Problem 17 [**]

Find the area of the circle.

[@cshearer41]

1. Use the cyclic quad to find the red square has a side length 8.
2. Then you can average the two chords to find the center of the circle and apply the Pythagorean theorem.

#### Problem 18 [**]

[@osdinato]

1. The key is symmetry. The center of top chord is length 2 and a perpendicular through it must run through the
center of the circle.

2. You can extend the lines to create 2 chords. By symmetry the horizontal one must be 8 in length and then you can use the cyclic quad to see the other piece length 4.

3. Finally apply the Pythagorean theorem on the right triangle created by the diameter and the 2 sides.

2. You can extend the lines to create 2 chords. By symmetry the horizontal one must be 8 in length and then you can use the cyclic quad to see the other piece length 4.

3. Finally apply the Pythagorean theorem on the right triangle created by the diameter and the 2 sides.

#### Problem 19 [***]

[@eylem_99]

This is one is quite tricky but you can find alot of interesting things while trying to solve it. First: note: this is a 36-72-72 and part of the pentagon. What follows is one Euclidean way to go:

1. Draw a parallel line to the outside right edge from the right angle to the left side.

2 Angle chase the 2 congruent triangles. formed.

3. Then from the parallel lines show they divide the right outside side exactly in half.

4. Find a isosceles triangle and from there it can be show the vertex is actually also half of the bottom angles.

## No comments:

## Post a Comment