Puzzles
Two triangles
The three sides of this triangle have been split into three equal parts and three lines have been added.
What is the area of the smaller blue triangle as a fraction of the area of the original large triangle?
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Draw on the following lines parallel to those which were added in the question.
Then a grid of copies of the smaller blue triangle has been created. Now consider the three triangles which are coloured green, purple and orange in the following diagram:
Each of these traingles covers half a parallelogram made from four blue triangles. Therefore the area of each of these triangles is twice the area of the small blue triangle.
And so the blue triangle covers one seventh of the large triangle.
Extension
If the sides of the triangle were split into \(n\) pieces the the lines added, what would the area of the smaller blue triangle be as a fraction of the area of the original large triangle?
Dodexagon
In the diagram, B, A, C, D, E, F, G, H, I, J, K and L are the vertices of a regular dodecagon and B, A, M, N, O and P are the vertices of a regular hexagon.
Show that A, M and E lie on a straight line.
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The interior angle of a regular hexagon is 120°. The interior angle of a regular dodecagon is 150°. Therefore angle CAM is 30°.
Now, consider the quadrilateral ACDE. This quadrilateral is symmetric (as the dodecagon is regular) so the angles CAE and DEA are equal. Hence:
$$360 = CAE+DEA+ACD+CDE\\
= 2CAE + 2\times 150\\
2CAE = 60\\
CAE=30
$$
The angles CAM and CAE are equal, so A, M and E lie on a straight line.
Extension
The vertices \(P_1\), \(P_2\), ..., \(P_n\) make up a regular \(n\)-gon and \(Q_1\), \(Q_2\), ..., \(Q_m\) make up a regular \(m\)-gon, with \(P_1=Q_1\) and \(P_2=Q_2\).
The vertices \(P_2\), \(Q_3\) and \(P_5\) lie on a straight line. What is the relationship between \(m\) and \(n\)?
Equal side and angle
In the diagram shown, the lengths \(AD = CD\) and the angles \(ABD=CBD\).
Prove that the lengths \(AB=BC\).
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By the sine rule in BCD:
$$\frac{CD}{\sin(CBD)}=\frac{BD}{\sin(BCD)}$$
By the sine rule in BAD:
$$\frac{AD}{\sin(ABD)}=\frac{BD}{\sin(BAD)}$$
\(ABD=CBD\) and \(AD=CD\), so:
$$\sin(BAD)=\frac{BD\sin(ABD)}{AD}\\
=\frac{BD\sin(CBD)}{CD}\\
=\sin(BCD)$$
Using the sine rule in ABC:
$$\frac{AB}{\sin(BCD)}=\frac{BC}{\sin(BAD)}$$
The two sines are equal and so:
$$AB=BC$$
Arctan
Prove that \(\arctan(1)+\arctan(2)+\arctan(3)=\pi\).
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Let \(\alpha=\arctan(1)\), \(\beta=\arctan(2)\) and \(\gamma=\arctan(3)\), then draw the angles as follows:
Extension
Can you find any other integers \(a\), \(b\) and \(c\) such that:
$$\arctan(a)+\arctan(b)+\arctan(c)=\pi$$
Square deal
This unit square is divided into four regions by a diagonal and a line that connects a vertex to the midpoint of an opposite side. What are the areas of the four regions?
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The square is unit, so \(a+b+c+d=1\). By the definitions of the lines, \(a+d=\frac{1}{2}\) and \(a+b=\frac{1}{4}\).
\(a\) and \(c\) are similar triangles. The vertical side of \(c\) is twice that of \(a\) so \(c=4a\).
Therefore we have the system of simultaneous equations:
$$a+b+c+d=1\\a+d=\frac{1}{2}\\a+b=\frac{1}{4}\\c=4a$$
These can be solved to find:
$$a=\frac{1}{12}\\b=\frac{1}{6}\\c=\frac{1}{3}\\d=\frac{5}{12}\\$$
Extension
What would be the areas if the lines were a diagonal and another line which divides the sides in the ratio \(x:y\)?
Half an equilateral triangle
What is the shortest straight line which bisects an equilateral triangle?
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Consider a line which bisects an equilateral triangle with sides of length 1. Let the sides of the smaller triangle created be \(a\) and \(b\) as shown below:
The area of the equilateral triangle is:
$$\frac{1}{2}\times 1\times 1\times \sin(60)\\=\frac{\sqrt{3}}{4}$$
The area of the smaller triangle is:
$$\frac{1}{2}ab \sin(60)\\=\frac{ab\sqrt{3}}{4}$$
This is half the area of the equilateral triangle, so:
$$ab=\frac{1}{2}$$
The length (\(L\)) of the line bisecting the equilateral triangle is:
$$L=a^2+b^2-2ab\cos(60)\\
=a^2+\left(\frac{1}{2a}\right)^2-\frac{2a}{2a}\times \frac{1}{2}\\
=a^2+\frac{1}{4a^2}-\frac{1}{2}$$
To find the minimum, we set the derivative to 0:
$$0=\frac{dL}{da}=2a-\frac{1}{2a^3}\\
4a^4-1=0\\
a^4=\frac{1}{4}\\
a=\frac{1}{\sqrt{2}}$$
For this value of \(a\), \(b\) and \(L\) also equal to \(\frac{1}{\sqrt{2}}\).
Extension
What if the line does not need to be straight?
Four points on a shape
Fiona wants to draw a 2-dimensional shape whose perimeter passes through the points A, B, C and D
Which of the following shapes can she draw?
(i) A circle
(ii) An equilateral triangle
(iii) A square
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(i) If a circle could be drawn, then its centre would be equidistant from each pair of points. But the locus of points equidistant from B and C is parallel to the locus of points equidistant from A and D.
Therefore it is impossible to place the centre of the circle, so no circle can be drawn.
(ii)
(iii)
Extension
Let A, B, C and D be any four points. When is it possible to draw
(i) A circle
(ii) An equilateral triangle
(iii) A square
through the four points?
Pizza
Twelve friends want to share a pizza. One of the friends is very fussy and will not eat the centre of the pizza.
Is it possible to split a (circular) pizza into twelve identical pieces such that there is at least one piece which does not touch the centre?
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![](javascript:showlimage("dodexagon.png"))
![](javascript:showlimage("equal-side-and-angle.png"))
![](javascript:showlimage("three-squares.png"))
![](javascript:showlimage("square-deal.png"))
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![](javascript:showlimage("four-points.png"))
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![](javascript:showlimage("four-points-square.png"))
![](javascript:showlimage("pizza-answer.png"))