Summer Problem Solving Marathon Question #25

[Value = 1 point]

The area of triangle ABC is 96. D is the midpoint of AB, E is the midpoint of DB, and F is the midpoint of BC. What is the area of the triangle AEF?

Summer Problem Solving Marathon Solution #23

Let D be the vertex of the rhombus on side AB of the triangle, E be the vertex of the rhombus on side AC, and F be the vertex of the rhombus on side BC.

Opposite sides of a rhombus are parallel, so DF is parallel to AE, and hence also to AC. So the triangles BDF and BAC are similar. Thus their sides are in the same ratio. Let s be the side length of the rhombus. Then BD = 12-s. The ratio of BD to BA is the same as the ratio of DF to AC. So we have:

(12 - s)/12 = s/6

Multiplying both sides by 12, we have:

12 - s = 2s
12 = 3s
s = 4

This gives us the side length of the rhombus, but not yet the area. To find the area of the rhombus, we view it as two congruent triangles -- ADE and FDE. To find the area of ADE, we use the area formula:

Area = (ab sin C)/2

In this case we have the two sides, so we just need the sine of the included angle. That angle is angle BAC.

We now use the law of cosines applied to triangle ABC. Recall that the law of cosines says:

c² = a² + b² - 2ab cos C

We use the particular instance in which c is side BC, a is side AB, b is side AC, and C is angle BAC (which we will call angle A). Then we have:

64 = 144 + 36 - 2(12)(6)cos A
64 = 180 - 144cos A
144cos A = 114
cos A = 114/144

However, we want sin A. So we use the fact that cos² A + sin² A = 1, so sin A = (1 - cos² A)^1/2.

So sin A = (1 - (114/144)²)^1/2 = ((144 - 114)²/144²)^1/2 = ((144 + 114)(144-114))^1/22/144 = (258 x 30)^1/2/144 = (6/144)√215 = (1/24)√215.

The area of triangle ADE is then (1/2)(4)(4)(1/24)√215 = (1/3)√215. The area of the rhombus is twice this, or (2/3)√215.

Summer Problem Solving Marathon Question #24

[Value = 9 points]

Let AOB be an acute angle, and Let C be a point on OA. From C, a perpendicular is drawn to OB, meeting OB at point D. The length of CD is a. From D, a perpendicular is drawn to OA, meeting OA at E. The length of DE is b. From E, a perpendicular is drawn to OB. From the base of this perpendicular, a perpendicular is drawn to OA. This procedure is repeated infinitely many times. What is the sum of the lengths of the perpendiculars dawn?

Summer Problem Solving Marathon Solution #22

In any quadratic equation with a leading coefficient of 1, the coefficient of x is the negative of the sum of the roots, and the constant is the product of the roots. So in our equation x² - ax + b = 0, the sum of the roots is a, and the product of the roots is b.

We are told that a, written in base n, is 18. Thus a = n + 8. Since n is one of the roots, the other root must be n.

The product of the roots is then b. So b = 8n. Written in base n, this is 80.

Summer Problem Solving Marathon Question #23

[Value = 4 points]

In triangle ABC, side AB is of length 12, side BC of length 8, and side AC of length 6. A rhombus is inscribed in ABC, with one vertex of the rhombus being at point A and two sides of the rhombus lying along AB and AC. What is the area of the rhombus?

Week Four Standings

Points through the end of the fourth week of the summer problem solving marathon:

Matthew: 122
Amelia: 62
Michael: 27
Sophia: 20
Paul: 3
Lily: 2
Ben: 2
Sharon: 2

Summer Problem Solving Marathon Question #22

[Value = 3 points]

The equation x² - ax + b = 0 has an integer solution n, where n is greater than 8. The coefficient a, written in base n, is 18. What is coefficient b written in base n?