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May's Problem

 

What is the smallest prime number that divides the following?
(13 × 17) + (19 × 23)

 


A solution to this problem will appear along with next month’s problem.

 

April's Problem

 

Points P, Q, and R lie on one line and points S, T, U, and V lie on another line as shown. What is the total number of different triangles that can be formed using any three of the points P, Q, R, S, T, U, and V as vertices?

 


Answer is:30

METHOD 1: Strategy: Count in an organized way.
One vertex is on one line and the other two are on the other line.

Suppose one vertex is on line 1 and two vertices are on line 2. There are 6 such triangles using vertex P: PST, PSU, PSV, PTU, PTV, and PUV. Likewise, there are 6 such triangles using vertex Q and 6 more using vertex R, for a total of 18.

Now suppose one vertex is on line 2 and two vertices are on line 1. There are 3 such triangles using vertex S: SPQ, SPR, and SQR. Likewise there are 3 such triangles using vertex T, 3 more using vertex U, and 3 more using vertex V, for a total of 12. In all, 30 triangles can be formed using any three of the points as vertices.

METHOD 2: Strategy (for more advanced students): Use combinatorics.
For each of the 3 points on line 1, there are 4C2 = 6 pairs of points on line 1. This gives us 18 triangles. For each of the 4 points on line 2, there are 3C2 = 3 pairs of points on line 1. This gives us 12 more triangles. Thus, a total of 30 triangles can be formed.

FOLLOW-UPS: (1) Suppose all 7 points are on a circle. How many triangles can be formed using any 3 of the points as vertices? [35] (2) See pages 111-114 in Creative Problem Solving in School Mathematics, 2nd Edition.

 

 

March's Problem

 

Rectangle ABCD is split into four smaller rectangles as shown. Each side of each rectangle is a whole number of cm. The areas of three of the small rectangles are shown. What is the area of rectangle ABCD, in sq cm?

 


Answer is: 184

METHOD: Strategy: Consider the possible dimensions of the small rectangles.
Rectangle AEJG, with area 21 sq cm, is either 1 cm by 21 cm, or 3 cm by 7 cm. Rectangle GJFD, with area 35 sq cm, is either 1 cm by 35 cm, or 5 cm by 7 cm. The common side, GJ, of both rectangles has a length of either 1 or 7 cm. If GJ = 1 cm, then AG = 21 cm. But 21 is not a factor of 48 and cannot be the length of a side of rectangle GJFD. Thus GJ = AE = DF = 7 cm and AG = EJ = BH = 3 cm. Then EB = JH = FC = 48 ÷ 3 = 16 cm, and GD = JF = HC = 5 cm. The area of rectangle JHCF = 16 × 5 = 80 sq cm, and the area of the large rectangle is 21 + 35 + 48 + 80 = 184 sq cm.

 

 

 

For many additional problems we highly recommend the following books:

Math Olympiad Contest Problems for Elementary and Middle Schools by Dr. G. Lenchner
and
Math Olympiad Contest Problems Volume 2 edited by Richard Kalman

are sources of many such problems.

Creative Problem Solving in School Mathematics 2nd Edition by Dr. George Lenchner

can help you to teach solving these types of problems