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Did you solve it? Are you in the smartest 1 per cent (of 13-year-olds)?

The solutions to today’s problems

Did you solve it? Are you in the smartest 1 per cent (of 13-year-olds)?
Photograph: Knauer/Johnston/Getty Images

Earlier today I set you the following puzzles:

1. In this word-sum, each letter stands for one of the digits 0–9, and stands for the same digit each time it appears. Different letters stand for different digits. No number starts with 0.

Find all the possible solutions of the word-sum shown above.

Solution

To find O, we add three Os to find a number that has an O in the units place. The only digits which satisfy this condition are 0 and 5.

If O = 0, there are no tens to “carry” to the middle column and so we are seeking a digit for M, for which three Ms have an M in the units place. Therefore, M is 0 or 5. However, O = 0 and so M = 5 (as different letters stand for different digits).

If O = 5, the total of three Os is 15 and so we have a ‘1’ to carry to the middle column. However, there are no digit values of M, where 3M + 1 gives M in the units column.

Hence, O = 0 and M = 5. If J = 1, I = 4. If J = 2, I = 7. If J > 2, 3 × ‘JMO’ does not give a 3-digit number and so there are only 2 possible values for J.

The only solutions are ‘JMO’ = 150 with ‘IMO’ = 450 and ‘JMO’ = 250 with ‘IMO’ = 750.

2. In the diagram below, a quarter circle with radius 3cm is positioned next to a quarter circle with radius 4cm.

What is the total shaded area bounded by the blue lines, in cm2.

Solution

Extend the bottom line to make a rectangle with height 1cm and width 3cm:

The area of a circle is πr2, where r is the radius, so the area bounded by the perimeter of the shape in the diagram above is the area of the 3x1 rectangle plus the two quarter circles:

3 + (1/4)9π + (1/4)16π = 3 + (25/4)π

The shaded area bound by the blue lines is this value minus the area of the long triangle of height 1cm and width (3 + 4)cm.

Which is 3 + (25/4)π – 7/2. So the answer is:

(25/4)π – (1/2).

3. An equilateral triangle is divided into smaller equilateral triangles.

The figure on the left shows that it is possible to divide it into 4 equilateral triangles. The figure on the right shows that it is possible to divide it into 13 equilateral triangles.

What are the integer values of n, where n > 1, for which it is possible to divide the triangle into n smaller equilateral triangles?

Solution

Let’s call the equilateral triangle which is to be divided into smaller equilateral triangles the original triangle.

Let’s start with n = 2. Is there a way of dividing the original triangle into two smaller ones? We can’t draw a line from a vertex, since this would produce angles less than 60°. But if we draw a single line that draws an equilateral triangle, as in (a) below, we create another quadrilateral in the original triangle, not a triangle. So n cannot be 2.

If we were going to be able to divide the quadrilateral in (a) into equilateral triangles, we could only divide it into an odd number of equilateral triangles, as in (b), which divides into 5 smaller equilateral triangles. Since there is no way to divide the quadrilateral into an even number of triangles, we can’t divide it into either 2 or 4 triangles, so we can also rule out n = 3 and n = 5.

The case for n = 4 is given in the question and so n = 4 is possible.

Diagram (b) shows the case for n = 6. Any of the smaller equilateral triangles in this diagram can be spilt into 4 equilateral triangles, which has the effect of adding 3 to the total number of triangles. (Diagram (c) is (b) with the top triangle divided into 4.) We can continue in this fashion in order to create cases when n = 9, 12, 15, . . . . Therefore, it is possible to split the original triangle into n smaller equilateral triangles where n is a multiple of 3, (excluding n = 3).

Returning to the diagram for n = 4, we can see that any equilateral triangle can be split into 4 equilateral triangles. Hence, n can be 7, 10, 13, . . . . Therefore, it it is possible to divide the original triangle into n smaller equilateral triangles where n is one more than a multiple of 3.

Diagram (d) shows the construction for splitting the original triangle into 8 smaller equilateral triangles. From this diagram, we can show that n = 11, 14, 17, . . . are possible by splitting any of the existing equilateral triangles into 4 equilateral triangles. Therefore, is is possible to split the original triangle into n smaller equilateral triangles where n is two more than a multiple of 3 (excluding n= 5.)

Therefore, is is possible to divide the triangle into n smaller equilateral triangles for all positive integer values of n, excluding 2, 3 and 5.

Thanks again to the United Kingdom Mathematics Trust for letting me use today’s problems, which come from this year’s Junior Mathematical Olympiad.

I’ll be back in two weeks.

I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.

Topic: #diagram
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