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June's Problem
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What year in the 21st century (the years 2001 to 2100) is the product of two consecutive integers?

 
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A solution to this problem will appear along with next month’s problem.
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May's Problem
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In triangle ABC, base AB is 10 centimeters longer than height CD. The sum of the base and height of the triangle is 26 centimeters. Square PQRS has twice the area of triangle ABC. What is the perimeter of square PQRS?

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METHOD 1: Use the average: The sum of base AB and height CD is 26, so their average is 13 cm. Since AB and CD differ from each other in length by 10 cm, each differs from their average by 5 cm. Then AB = 13 + 5 = 18 cm and CD = 13 - 5 = 8 cm. The area of the triangle is ½ × 18 × 8 = 72 sq cm. Then the area of the square is 144 sq cm, the length of any one side is 12 cm and the perimeter of the square is 48 cm.

METHOD 2: Make a table:
To find the lengths of AB and CD, create a table of number pairs whose sum is 26. The only pair whose difference is 10 is 18 and 8. (Or create a table of number pairs whose difference is 10 and see which pair also has a sum of 26.) Then proceed as above to find the area of the triangle, followed by the area, side-length, and perimeter of the square.

METHOD 3: Solve a simpler problem:
Since AM measures 10 more than CD, subtracting 10 from AM gives equal lengths. Subtracting 10 also from the sum would be twice the length of CD. 26 - 10 = 16 and 16/2 is 8. Thus CD = 8 and AB = 8 + 10 = 18 cm. Proceed as above to find the area of the triangle, followed by the area, side-length, and perimeter of the square.

METHOD 4: Use algebra:
Let x represent the height CD, then x + 10 represents the length of the base AB. We are given that x + (x + 10) = 26. Solving the equation, we have x = 8. Proceed as above.
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April's Problem
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If the lengths of the three sides of a triangle are added two at a time, the sums are 20, 23, and 27. What is the perimeter of the triangle?

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METHOD 1 Find all three sides: Because a + b = 20 and a + c = 23, c is 3 more than b. Compare that to b + c = 27; it follows that b = 12 and c = 15. Since a + b = 20, a = 8. (Check your answers in all three equations.) Then the perimeter of the triangle is 8 + 12 + 15 = 35 .

METHOD 2 Add all equations:
To find the perimeter, find the value of a + b+ c. First add the three equations: the result is (a + b) + (a + c) + (b + c) = 20 + 23+ 27. This simplifies to 2a + 2b + 2c = 70. Divide by 2 to get a + b+ c = 35.

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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
and
MOEMS® Contest Problems Volume 3
edited by Richard Kalman & Nicholas J. Restivo.
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