Puzzles
19 December
The diagram to the right shows a triangle. Two of the sides of the triangle have been split into three pieces, with lines drawn from the opposite vertex. In total, the diagram now contains 27 triangles of any size.
Another triangle has two of its sides split into eight pieces, with lines drawn from the opposite vertex. How many triangles (of any size) would this create?
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Each triangle must include one of the two corners at the base of the largest triangle (or both those corners).
To make triangles including the bottom left corner of the large triangle, we must pick two lines coming out of that corner, and one line coming out of the bottom right corner that is not the base of the triangle.
There are 9 lines coming out of the corners, so the total number of triangles is \(\left(\begin{array}{c}9\\2\end{array}\right)\times\left(\begin{array}{c}8\\1\end{array}\right)=288\).
We now need to count the triangles that include the bottom right corner of the large triangle, but do not include both corners (as we've already counted thoses).
There are \(\left(\begin{array}{c}8\\2\end{array}\right)\times\left(\begin{array}{c}8\\1\end{array}\right)=224\) ways to pick lines to make these triangles.
In total, this makes 512 triangles.
5 December
28 points are spaced equally around the circumference of a circle. There are 3276 ways to pick three of these points.
The three picked points can be connected to form a triangle. Today's number is the number of these triangles that are isosceles.
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Pick one of the 28 points and imagine a line to the point opposite it. There will be 13 points on each side of this line. Picking one of these 13 points and then picking the
corresponding point on the other side of the line gives an isosceles triangle. Therefore there are 13 isosceles triangles with the first chosen point as the point where the two equal sides meet.
There were 28 choices for the first point, and so the total number of isosceles triangles will be 13×28=364.
2 December
You have 15 sticks of length 1cm, 2cm, ..., 15cm (one of each length). How many triangles can you make by picking three sticks and joining their ends?
Note: Three sticks (eg 1, 2 and 3) lying on top of each other does not count as a triangle.
Note: Rotations and reflections are counted as the same triangle.
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In order to make a valid triangle, the sum of the two shorter sides must be larger than the longest side.
There is
1 triangle with longest side 4cm (2,3,4);
2 with longest side 5cm (2,4,5 and 3,4,5);
4 with longest side 6cm (2,5,6; 3,5,6; 4,5,6 and 3,4,6);
6 with longest side 7cm (2,6,7; 3,6,7; 4,6,7; 5,6,7; 3,5,7 and 4,5,7);
9 with longest side 8cm;
12 with longest side 9cm;
16 with longest side 10cm;
20 with longest side 11cm;
25 with longest side 12cm;
30 with longest side 13cm;
36 with longest side 14cm; and
42 with longest side 15cm.
In total this makes 203 triangles.
12 December
There are 2600 different ways to pick three vertices of a regular 26-sided shape. Sometimes the three vertices you pick form a right angled triangle.
These three vertices form a right angled triangle.
Today's number is the number of different ways to pick three vertices of a regular 26-sided shape so that the three vertices make a right angled triangle.
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The vertices of the 26-gon lie on a circle. The triangle is therefore right-angled if (and only if) the longest side is a diameter of the circle.
In other words, the triangle is right angled if (and only if) two of its vertices are opposite vertices of the 26-gon.
There are 13 different pairs of opposite points on the 26-gon. For each of these, there are 24 remaining vertices that could be the third vertex of the triangle.
Therefore there are 13×24=312 different right angled triangles.
Is it equilateral?
In the diagram below, \(ABDC\) is a square. Angles \(ACE\) and \(BDE\) are both 75°.
Is triangle \(ABE\) equilateral? Why/why not?
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The triangle is equilateral.
To see why, add a copy of point \(E\) rotated by 90°. This is labelled \(F\) on the diagram below.
Angles \(BDE\) and \(CDF\) are both 75°. Therefore angles \(CDE\) and \(BDF\) are both 15°. This means that angle \(FDE\) is 60°.
Line \(AD\) is a line of symmetry of the diagram, so angles \(DFE\) and \(DEF\) are equal and both 60°. Therefore, triangle DEF is equilateral. This triangle is show in green in the diagram below.
Lines \(EF\), \(DF\) and \(BF\) are all equal length, so triangles \(BFE\) and \(BFD\) are isosceles.
Angles \(BDF\) and \(FBD\) are both 15°. Angles \(FBE\) and \(FEB\) are equal, and the angles in triangle \(BED\) add to 180°: this means that angle \(FBE\) is 15°.
Angles \(FBE\) and \(FBD\) are both 15°, and so angle \(EBD\) is 30°. Angles \(EBD\) and \(ABE\) add to 90°, and so angle \(ABE\) is 60°.
By symmetry, angle \(BAE\) is also 60°. Angle \(BEA\) must therefore also be 60°, so triangle \(ABE\) is equilateral.
20 December
Start with a line of length 2. Draw a line of length 17 perpendicular to it. Connect the ends to make a right-angled triangle.
The length of the hypotenuse of this triangle will be a non-integer.
Draw a line of length 17 perpendicular to the hypotenuse and make another right-angled triangle. Again the new hypotenuse will have a non-integer length.
Repeat this until you get a hypotenuse of integer length. What is the length of this hypotenuse?
5 December
How many different triangles are there with a perimeter of 100 and each side
having an integer length?
(different = not rotations or reflections)
Cutting corners
The diagram below shows a triangle \(ABC\). The line \(CE\) is perpendicular to \(AB\) and the line \(AD\) is perpedicular to \(BC\).
The side \(AC\) is 6.5cm long and the lines \(CE\) and \(AD\) are 5.6cm and 6.0cm respectively.
How long are the other two sides of the triangle?
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Let \(A\), \(B\) and \(C\) represent the angles at points \(A\), \(B\) and \(C\). Looking at triangle \(ACE\) gives:
$$\sin A=\frac{5.6}{6.5}$$
And triangle \(ACD\) gives:
$$\sin C=\frac{6.0}{6.5}$$
Using Pythagoras' Theorem to find the missing sides in these triangles also gives:
$$\cos A=\frac{3.3}{6.5}$$
$$\cos C=\frac{2.5}{6.5}$$
We can use these to find \(\sin B\):
$$\begin{array}{rl}
\sin B&=\sin(\pi-A-C)\\
&=\sin(A+B)\\
&=\sin A\cos C + \cos A\sin C\\
&=\frac{5.6}{6.5}\frac{2.5}{6.5}+\frac{6.0}{6.5}\frac{3.3}{6.5}\\
&=\frac{33.8}{6.5^2}
\end{array}$$
Now the missing sides can be found using the sine rule:
$$\begin{array}{rl}
CB&=\frac{CA\sin A}{\sin B}\\
&=\frac{6.5\times\frac{5.6}{6.5}}{\frac{33.8}{6.5^2}}\\
&=\frac{6.5^2\times5.6}{33.8}\\
&=7\text{cm}
\end{array}$$
$$\begin{array}{rl}
AB&=\frac{CA\sin C}{\sin B}\\
&=\frac{6.5\times\frac{6}{6.5}}{\frac{33.8}{6.5^2}}\\
&=\frac{6.5^2\times6}{33.8}\\
&=7.5\text{cm}
\end{array}$$
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