Category Archives: Geometry

Problem Solving

I am came across this problem and was fascinated. It’s from this book

At first I went straight to the 14-sided polygon, and tried to draw the diamonds (parallelograms), but then I thought let’s start smaller and see if there is a pattern.

Clearly a square contains 1 diamond (itself).

Pentagon

It’s not possible with a pentagon.

Hexagon

A hexagon has 6 diamonds

Septagon

I am guessing it’s not possible to fill a regular 7-sided shape with diamonds

It’s not possible with odd numbers of sides. Regular polygons with an odd number of sides have no parallel sides, so we can’t cover it with rhombi (which have opposite sides parallel).

Octagon

An octagon has 6 diamonds.

We know a decoagon has 10 diamonds (from the question)

Let’s put together what we know

n46810
Diamonds13610

These are the triangular numbers, so when n=12 the number of diamonds is 15, and for n=14 it’s 21.

We can work out a rule for calculating the number of diamonds given the number of sides.

Because the difference in the n values is not 1, I am going to get n and D in terms of k and then combine the two equations.

From the above table, n=2k+2

We know this rule is quadratic as the second difference is constant, hence

D=\frac{1}{2}k^2+bk+c

    \begin{equation*}1=\frac{1}{2}+b+c\end{equation}

(1)   \begin{equation*}\frac{1}{2}=b+c\end{equation*}

    \begin{equation*}3=\frac{1}{2}2^2+2b+c\end{equation}

(2)   \begin{equation*}1=2b+c\end{equation*}

Solve simultaneously, subtract equation 1 from equation 2

(3)   \begin{equation*}\frac{1}{2}=b\end{equation*}

Substitute for b=\frac{1}{2} into equation 1

    \begin{equation*}\frac{1}{2}=\frac{1}{2}+c\end{equation}

c=0, therefore D=\frac{1}{2}k^2+\frac{1}{2}k

We know n=2k+2 hence k=\frac{n-2}{2}

Hence D=\frac{1}{2}(\frac{n-2}{2})^2+\frac{1}{2}(\frac{n-2}{2})

D=\frac{1}{8}(n^2-4n+4)+\frac{1}{4}(n-2)=\frac{n^2}{8}-\frac{n}{2}+\frac{1}{2}+\frac{n}{4}-\frac{1}{2}=\frac{n^2}{8}-\frac{n}{4}

    \begin{equation*}D=\frac{n^2}{8}-\frac{n}{4}\end{equation}

Let’s test our rule for n=14

    \begin{equation*}D=\frac{14^2}{8}-\frac{14}{4}=\frac{49}{2}-\frac{7}{2}=\frac{42}{2}=21\end{equation}

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Filed under Area, Geometry, Interesting Mathematics, Puzzles, Quadratics

Area/Geometry Problem

This problem is from The Geometry Forum Problem of the Week June 1996

In triangle ABC, AC=18 and D is the point on AC for which AD=5. Perpendiculars drawn from D to AB and CB have lengths of 4 and 5 respectively. What is the area of triangle ABC?

I put together a diagram (in Geogebra)

Add points P and Q

Triangle APD and triangle DQC are right angled. Using pythagoras, AP=3 and QC=12

BQDP is a cyclic quadrilateral and BD is the diameter. I am not sure if this is useful, but it is good to notice.

    \begin{equation*}sin(A+B+C)=sin(180)=0\end{equation}

    \begin{equation*}sin((A+C)+B)=sin(A+C)cosB+sinBcos(A+C)=0\end{equation}

    \begin{equation*}cosB(sinAcosC+sinCcosA)+sinB(cosAcosC-sinAsinC)=0\end{equation}

    \begin{equation*}cosB(\frac{4}{5}\times\frac{12}{13}+\frac{5}{13}\times\frac{3}{5})+sinB(\frac{3}{5}\times\frac{12}{13}-\frac{4}{5}\times\frac{5}{13})=0\end{equation}

    \begin{equation*}cosB(\frac{48}{65}+\frac{15}{65})+sinB(\frac{36}{65}-\frac{20}{65})=0\end{equation}

    \begin{equation*}\frac{63}{65}cosB+\frac{16}{65}sinB=0\end{equation}

    \begin{equation*}63cosB+16sinB=0\end{equation}

    \begin{equation*}63+16tanB=0\end{equation}

    \begin{equation*}tanB=\frac{-63}{16}\end{equation}

If tanB=\frac{-63}{16} then sinB=\frac{63}{65}

Now,

    \begin{equation*}\frac{y+12}{sinA}=\frac{18}{sinB}\end{equation}

    \begin{equation*}y+12=\frac{4}{5}(18)\frac{65}{63}\end{equation}

    \begin{equation*}y+12=\frac{104}{7}\end{equation}

Hence the Area is

    \begin{equation*}A=\frac{1}{2}(18)(\frac{104}{7})sinC\end{equation}

    \begin{equation*}A=\frac{1}{2}(18)(\frac{104}{7})\frac{5}{13}\end{equation}

    \begin{equation*}A=\frac{360}{7}=51.43\end{equation}

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Filed under Area, Finding an area, Geometry, Identities, Non-Right Trigonometry, Pythagoras, Trigonometry

Using a Vector Method to Find an Angle Bisector

Points A and B are defined by the position vectors \mathbf{a}=3\mathbf{i}+4\mathbf{j} and \mathbf{b}=12\mathbf{i}+5\mathbf{j}.

Find a vector that bisects \angle{AOB}.

If we think about how we add vectors using the parallelogram rule

Finding the resultant vector using the parallelogram rule

we can take advantage of the geometric properties of parallelograms (or of a rhombus).

If \mathbf{a} and \mathbf{b} are unit vectors, then the parallelogram is a rhombus, and the diagonal (i.e the resultant) bisects the angle.

We need to find the sum of the unit vectors.

|\mathbf{a}|=\sqrt{3^2+4^2}=5

\therefore \hat{\mathbf{a}}=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}

|\mathbf{b}|=\sqrt{12^2+5^2}=13

\therefore \hat{\mathbf{b}}=\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}

The vector that bisects \angle{AOB} is

\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}+\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}=\frac{99}{65}\mathbf{i}+\frac{64}{65}\mathbf{j}

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Filed under Geometry, Vectors, Year 11 Specialist Mathematics

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Find the area of an n-sided regular polygon if you know the side length, l.

An octagon for a visual reference

Find the h of the triangle in terms of l and theta.

tan(\theta)=\frac{\frac{l}{2}}{h}

h=\frac{l}{2tan(\theta)}

Remember the area of a triangle is A=\frac{1}{2}bh

Hence, A=\frac{1}{2} l \times \frac{l}{2tan(\theta)}=\frac{l^2}{4tan(\theta)}

And \theta=\frac{360}{2n}=\frac{180}{n}

Therefore A=\frac{l^2}{4tan(\frac{180}{n})}

There are n triangles in an n-sided polygon

(1)   \begin{equation*}A=\frac{nl^2}{4tan(\frac{180}{n})}\end{equation*}

Find the area of a hexagon with side length 10cm.
A=\frac{6\times10^2}{4tan(\frac{180}{6})}
A=\frac{600}{4(\frac{1}{\sqrt{3}})}
A=150\sqrt{3} cm^2

Finding the area of a polygon if you know the inradius or the apothem

The apothem and the inradius are the same. It is the radius of the incircle.

Find the area of the triangle in terms of a and theta.

tan(\theta)=\frac{\frac{l}{2}}{a}

l=2atan(\theta)

A=\frac{1}{2}bh

A=\frac{1}{2}2atan(\theta)a=a^2tan(\theta)

And \theta=\frac{180}{n}

Hence for an n-sided polygon

(2)   \begin{equation*}A=na^2tan(\frac{180}{n})\end{equation*}

Find the area of a regular pentagon with apothem 4.5cm
A=5\times 4.5^2tan(\frac{180}{5})
A=73.56cm^2

Finding the area of a regular polygon given the circumradius

The circumradius is the radius of the circumscribed circle (R in the diagram above)

Remember the area of triangle formula

A=\frac{1}{2}absin(\theta)

A=\frac{1}{2}R^2sin(\theta)

\theta=\frac{360}{n}

Hence, A=\frac{1}{2}R^2sin(\frac{360}{n})

Hence, for an n-sided polygon

(3)   \begin{equation*}A=\frac{nR^2sin(\frac{360}{n})}{2}\end{equation*}

Find the area of a regular octagon inscribed in a circle of radius 10cm.
A=\frac{8\times 10^2sin(45)}{2}
A=200\sqrt{2}cm^2

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Filed under Area, Area of Triangles (Sine), Finding an area, Non-Right Trigonometry, Regular Polygons, Right Trigonometry, Year 11 Mathematical Methods

Intersecting Secant Theorem

CD is a tangent to the circle.

Prove c^2=a(a+b)

I am going to add two chords to the circle

Chord AD and BD are added

\angle{BDC}=\angle{CAD} (angles in alternate segments are congruent)

\angle{BCD}=\angle{DCA} (shared angle)

\therefore \Delta BDC\cong \Delta{DAC} (AA)

Hence

\frac{DC}{AC}=\frac{BC}{DC} (Corresponding sides in similar triangles)

\frac{c}{a+b}=\frac{a}{c}

\therefore c^2=a(a+b)

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Filed under Circle Theorems, Geometry, Year 11 Specialist Mathematics

Circle Geometry Question 2

One of my Year 11 Specialist students had this question

Triangle ABC touches the given circle at Points P, Q, R and S only. The secant BW touches the circle at V and W.

Diagram not drawn to scale

(a) Determine the lengths of the line segments marked x, y and z, leaving your answers as exact values.

(b) If the length of the line segment QW is 4 units, determine the exact radius of the circle.

(a) We are going to use the Intersecting Secant Theorem – the tangent version

c^2=a\times(a+b)

Hence, we have

    \begin{equation*}30^2=25(25+x+6)\end{equation}

    \begin{equation*}900=25(31+x)\end{equation}

    \begin{equation*}x=5\end{equation}

Then we can use the intersecting chord theorem to find y.

    \begin{equation*}10\times y=6 \times x\end{equation}

    \begin{equation*}10y=30\end{equation}

    \begin{equation*}y=3\end{equation}

Back to the Intersecting Secant Theorem to find z

    \begin{equation*}z^2=4\times 17\end{equation}

    \begin{equation*}z=2\sqrt{17}\end{equation}

(b)


QW is part of a 3-4-5 triangle, therefore \angle{Q}=90^\circ

This is definitely the case of the diagram not being drawn to scale. If \angle{Q}=90^\circ, then the purple line must be the diameter.

We can use pythagoras to find the length of the diameter

    \begin{equation*}(2r)^2=13^2+4^2\end{equation}

    \begin{equation*}4r^2=185\end{equation}

    \begin{equation*}r=\frac{\sqrt{285}}{2}\end{equation}

The radius of the circle is \frac{\sqrt{285}}{2}

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Filed under Circle Theorems, Geometry, Pythagoras, Year 11 Specialist Mathematics

Circle Geometry Question

In the above diagram O is the centre of the larger circle. A, B,D and E are points on the circumference of the larger circle. A, C, E and 0 are points on the circumference of the smaller circle. Show that \angle{CAB}=\angle{ABC}. AB, AC and BC are straight lines.

AO=OB (radii of the larger circle)

At a line from O to E (it is also a radius of the larger circle)

Let \angle{CAB}=\alpha.

ACEO is a cyclic quadrilateral.

Hence, \angle{CED}=180-\alpha (AECO is a cyclic quadrilateral)

As CB is a straight line \angle{OEB}=180-(180-\alpha)=\alpha.

\Delta OEB is an isosceles triangle.

Therefore, \angle{ABC}=\alpha

Therefore \angle{ABC}=\angle{CAB}

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Filed under Circle Theorems, Finding an angle, Geometry, Year 11 Specialist Mathematics

Tricky Area Ratio Question

A circle of radius r, is inscribed within an isosceles triangle ABC. CA=CB=5r. Given that \angle{ACB} is acute, find the ratio of the area of the circle to that of triangle ABC.

Mathematics Specialist 3AB Question15 page 55

I came across this question while searching for an area of sectors and segments question.

Here’s a diagram

We know AC, AB and BC are tangents to the circle. Because the triangle is isosceles, the distance from A to the circle is the same as the distance from B to the circle.

CP is perpendicular to AB (because the triangle is isosceles).

Because it is proportional, i.e. r and 5r, we can let r=1

Let h=CP

(1)   \begin{equation*}h=\sqrt{25-a^2}\end{equation*}

but h also equals

(2)   \begin{equation*}h=1+\sqrt{(5-a)^2+1}\end{equation*}

Set equation 1 equal to equation 2

    \begin{equation*}\sqrt{25-a^2}={1+\sqrt{(5-a)^2+1}\end{equation}

Square both sides

    \begin{equation*}25-a^2=1+(5-a)^2+1+2\sqrt{(5-a)^2+1}\end{equation}

Expand and simplify

    \begin{equation*}25-a^2=2+25-10a+a^2+2\sqrt{(5-a)^2+1}\end{equation}

    \begin{equation*}0=2-10a+2a^2+2\sqrt{(5-a)^2+1}\end{equation}

Divide by 2

    \begin{equation*}0=1-5a+a^2+\sqrt{(5-a)^2+1}\end{equation}

    \begin{equation*}-\sqrt{(5-a)^2+1}=a^2-5a+1\end{equation}

Square both sides

    \begin{equation*}(5-a)^2+1=a^4-5a^3+a^2-5a^3+25a^2-5a+a^2-5a+1\end{equation}

    \begin{equation*}25-10a+a^2+1=a^4-10a^3+27a^2-10a+1\end{equation}

    \begin{equation*}0=a^4-10a^3+26a^2-25 \end{equation}

I solved this using a graphics calculator

a=-0.8434, a=1.3068, a=4.5367, a=5

We can reject a=-0.8434, a=4.5367, and a=5. If a=5, there isn’t a triangle, and if a=4.5367 \angle{ACB} is not acute.

Hence the area of triangle ABC=a\times h

    \begin{equation*}A_{\Delta}=1.3068\times\sqrt{25-1.3068^2}\end{equation}

(3)   \begin{equation*}A_{\Delta}=6.3069\end{equation*}

(4)   \begin{equation*}A_{\circ}=\pi\end{equation*}

Hence, equation 4 divided by equation 3 is

    \begin{equation*}\frac{\pi}{6.3069}\approx 0.5\end{equation}

Perhaps I approached this question in the wrong way. Is there an easier process?

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Filed under Area, Geometry, Pythagoras

Intersecting Circles

Two circles of radius L and 2L intersect as shown. What is the area of the shaded region?

From Professor Povey’s Perplexing Problems

My plan is to find the sum of the area of the two segments (see below).

Construct triangles

The diagonal of the square (the pink line above) has length

=\sqrt{(2L)^2+(2L)^2}=\sqrt{8L^2}=2\sqrt{2}L

From the pink triangle in the above diagram, I am going to find the angles using the cosine rule.

cos\theta=\frac{L^2+(2\sqrt{2}L)^2-(2L)^2}{2(L)(2\sqrt{2}L)}

cos\theta=\frac{5L^2}{4\sqrt{2}L^2}

cos\theta=\frac{5}{4\sqrt{2}}

\theta=0.487

cos\alpha=\frac{(2L)^2+(2\sqrt{2}L)^2-L^2}{(2\sqrt{2}L)(2L)}

cos\alpha=\frac{11L^2}{8\sqrt{2}L^2}

cos\alpha=\frac{11}{8\sqrt{2}}

\alpha=0.236

The green quadrilateral is a kite, which means the diagonals are perpendicular.

This means the segment angles are 2\theta and 2\alpha (because the triangles are isosceles and the diagonal is perpendicular to the base of the triangles).

Area of green segment

A=\frac{1}{2}r^2(\theta-sin\theta)

A=\frac{1}{2}L^2(0.9734-sin(0.9734))=0.0733L^2

Area of yellow segment

A=\frac{1}{2}4L^2(0.4721-sin(0.4721))=0.0035

Total Area =0.0733L^2+0.0035L^2=0.108L^2

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Filed under Finding an area, Geometry, Professor Povey

Geometry Problem

This problem is from Geometry Snacks by Ed Southall and Vincent Pantaloni – it’s a great book.

Two squares are constructed such that three vertices are collinear as shown. Find the value of the marked angle.

I started by marking in the right angles. And I added the diagonal of the larger square (pink line).

Because there are right angles at O and P, we know there is a circle, which has the diagonal of the square as its diameter (see second image below).

\angle{RSP} is 45^{\circ} (Angle between the diagonal and side of a square)

PORS is a cyclic quadrilateral.

In cyclic quadrilaterals opposite angles are supplementary.

Hence, \angle{ROP}=180^{\circ}-45^{\circ}=135^{\circ}

As \angle{ROS}=90^{\circ}, \angle{SOP} must be 45^{\circ}

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Filed under Finding an angle, Geometry