This post covers a second set of geometry problems from the Intermediate Maclaurin Mathematical Olympiad (a mathematics competition run by the UK Mathematics Trust, aimed at students of age 16). Like the problems in the previous post, these problems do not require knowledge beyond that of a GCSE-level student, they are more a test of one’s creativity and insight than a test of one’s knowledge of the subject.
The two key pieces of information in this problem’s description are that all three shapes are regular, and that they have an edge in common. Another way of phrasing this is that every visible line in the question’s diagram has the same length. Additionally, since all three shapes are regular, they all have their own sets of equal internal angles. We can calculate the size of a single internal angle from all three shapes using the formula below (where n is the number of sides the polygon has):
Sum of the interior angles of a polygon (degrees) = (n-2) * 180
Single internal angle (degrees) = Sum of the interior angles of a polygon (degrees) / n
Using this procedure, we acquire the following sizes for a single internal angle of each polygon:
- 15-gon: 156 degrees
- Heptagon: 900/7 degrees
- Decagon: 144 degrees
In order to find the size of angle XYZ, we need to take into account two ideas: one being that the sum of the angles around a single point is 360 degrees, and another being that the sum of the angles inside a triangle is 180 degrees. To progress further towards a solution, we can first focus on the lower point of the side the three polygons have in common (this point will be named P). We can find the size of the angle XPY by subtracting 900/7 degrees and 144 degrees from 360 – doing this gives the value 612/7. Additionally, we can also find the size of the angle ZPY by subtracting 900/7 degrees from 156 degrees – doing this gives the value 192/7. Given that the lengths of the lines PZ and PY are equal (since every visible line in the question’s diagram has the same length), we can state that the triangle ZPY is an isosceles triangle, and we can find the angle PYZ by subtracting 192/7 degrees from 180 degrees and dividing the result by 2 – this gives the value 534/7. To find the size of the angle XYZ, we can subtract the size of the angle XYP from the size of the angle PYZ. Getting the size of the angle XYP can be done by subtracting the size of the angle XPY from 180, and dividing the result by 2. To find the size of the angle XPY, we can subtract the values 144 degrees and 900/7 degrees from 360 degrees to obtain the value 612/7. Subtracting this value from 180 degrees and dividing the result by 2 gives the value 324/7 for the size of the angle XYP. We can now find the size of the angle XYZ by subtracting 324/7 from 534/7; doing this gives the value 210/7, which simplifies to 30 degrees.
In order to tackle this question, we must be aware of the circle theorem which states that a triangle drawn inside a semicircle with the longest side being said semicircle’s diameter will always be a right-angled triangle. Therefore, the triangle formed by the three points C, A and D is a right-angled triangle. Given that we know that the length of this right-angled triangle’s hypotenuse is 4, we can use the Pythagoras Theorem to find the length of CD if we know the distance between the two points A and C.
A kite can be drawn inside the circle if a radius is drawn to the point C. A key characteristic of this kite is that the length from its lowest point to its highest point is 2 (because this length is the radius of the circle).
This kite can then be split into four right-angled triangles, with the side lengths shown in the image below:
In the drawing above, we acknowledge that the distance from the lowest point of the kite to the highest point is 2, and we split this distance at the line AC to form two lines of length x and 2-x. Additionally, we know that the shape above is symmetrical, so the line from its lowest point to its highest point bisects the line AC into two smaller lines of length y. Focusing on one of the two sides of the kite, we can use the Pythagoras Theorem to form the following equations:
First we observe that x^2 and y^2 sum to 1, and then we can subtract this expression from the sum of y^2 and (2-x)^2 to obtain a linear equation which we can then solve to find the value of x. We can then substitute this value of x into our first equation in order to acquire the positive value of y (obviously, a negative value of y does exist, but here we disregard it because we are dealing with length).
Now we can go back to our original image of the circle – given that the line AC has a length of 2y, we can apply the Pythagorean Theorem as follows to find the length of CD (bearing in mind that the circle theorem mentioned at the start of the explanation for this problem states that the triangle ACD is right-angled).
This problem initially appears quite complex – it requires a crucial insight in order for it to be solved, however, with this insight, the solution to the problem is arguably the most concise and elegant out of the three solutions on this post. The various regions in the diagram are created by the intersections of two triangles with equal area, shown below:
We can see that these two triangles have equal area by looking at the formula below:
Area of a triangle = (base * perpendicular height) / 2
Given that the length of the base of one of the triangles is the perpendicular height of the other triangle, and that the multiplication operation is commutative, this formula returns the same value for the area of both triangles (we do not need to know any specific values for either the base or the height in order to see that this is the case). Additionally, this value for the areas of the triangles is equal to half the area of the entire diagram, because the triangles share their dimensions with the rectangle, and therefore the white area on the right diagram is equal to the shaded area on the left diagram – if we label the regions on the diagram in the manner shown below, we can form and solve a new equation based on this idea.
Looking at the two previous diagrams, the shaded area of the one on the left contains the regions a, b and c, while the white area of the one on the right contains the regions of a and c, along with three regions of sizes 1, 2 and 3. Given that both sections have equal area, we can form and solve the following equation to find the area of region b, which is the shaded area in the original question’s diagram.
As the Maclaurin Olympiad draws closer, I will continue to make more posts on the problems it has given in the past. Additionally, I will also begin writing about an interesting computing-related side project of mine in the coming weeks – more about this project will be revealed soon.