PG 25 4 Apr 2019Easter challenges

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Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

  1. a portmanteaux word combining parallel and telegram. A message sent each week by the Parallel Project to bright young mathematicians.
  • Tackle each Parallelogram in one go. Don’t get distracted.
  • Finish before the end of Easter holidays.
  • Your score & answer sheet will appear immediately after you hit SUBMIT.
  • Don’t worry if you score less than 50%, because it means you will learn something new when you check the solutions.

As it is the Easter holidays, you have a bit more time to complete this Parallelogram, which is why it is twice as long as a typical Parallelogram. It includes eight Intermediate Maths Challenge problems. The first four are tricky, but the second four are downright nasty, so give them some serious thought and see if you can solve them.

Because this Parallelogram is longer and tougher than a typical Parallelogram, we will be awarding TREBLE POINTS for this week's answers! So there's even more reason to give this one your best shot.

1. Venn diagrams

This is my all time favourite Venn diagram. What? Doesn’t everyone have a favourite Venn Diagram? Anyway... If you don’t yet have a favourite Venn diagram, then you can borrow this one for the time being. Can you work out the label for the two circles? You can find my answer below.

I guess that the left set contains “creatures with flat tails playing guitars”, while the right set contains “creatures with bills playing keyboards”. Hence, the animal in the overlap is a platypus playing a keytar, because it is a creature with a bill AND a flat tail, which is playing a guitar AND a keyboard.

2.

3 marks

2.1 Jane has 20 identical cards in the shape of an isosceles right-angled triangle. She uses the cards to make the five shapes below.

Which of the shapes has the shortest perimeter?

  • A
  • B
  • C
  • D
  • E
Show Hint (–1 mark)
–1 mark

The isosceles triangle is also a right-angle triangle, so it has one long side, and two short sides. So one option is to measure the perimeter of each shape in terms of shorts and longs. So, shape (D) is 6 shorts.

Let the side lengths of the isosceles right-angled triangle be x, x and y as shown in the diagram.

By Pythagoras’ Theorem y2=x2+x2=2x2 and hence y=2x.

We then see that the perimeters of the shapes are as follows:

  • A: 42x
  • B: 4+22x
  • C: 4+22x
  • D: 6
  • E: 4+22x.

Now, as 1<2<1.5, we have 42<4×1.5=6=4+2<4+22. So shape A has the shortest perimeter.

[Note: that, as the value of x is irrelevant, we could have, for example, put x=1 to begin with.]

3.

3 marks

3.1 ABCDE is a regular pentagon and BCF is an equilateral triangle such that F is inside ABCDE. What is the size of FAB?

  • 48°
  • 63°
  • 66°
  • 69°
  • 72°
Show Hint (–1 mark)
–1 mark

AB=FB, so the triangle ABF is isosceles, and hence FAB=BFA.

Since the pentagon ABCDE is regular, BC=AB. Since the triangle BCF is equilateral, BC=FB. Therefore, AB=FB. So the triangle ABF is isosceles, and hence FAB=BFA.

ABC=108°, as it is the interior angle of a regular pentagon. As the triangle BCF is equilateral, FBC=60°. Therefore ABF=ABCFBC=108°60°=48°.

So, from triangle ABF, we can deduce that FAB+BFA=180°48°=132°.

Since FAB=BFA, it follows that FAB=12132°=66°.

4.

3 marks

4.1 For which of the following numbers is the sum of all its factors not equal to a square number?

  • 3
  • 22
  • 40
  • 66
  • 70

The most straightforward method here is to just list all the factors of the numbers, add them up, and see in which cases this sum is not a square number:

Since 90 is not a square, in the context of the IMC, you could stop here. However, for completeness, we also calculate the sum of the factors of the other two numbers we are given.

So, for each option given, other than 40, the sum of its factors is a square number.

5.

3 marks

5.1 The sum:

one+four=seventy

becomes correct if we replace each word by the number of letters in it to give 3 + 4 = 7.

Using the same convention, which of these words could be substituted for x to make the following sum true?

three+five=x

  • eight
  • nine
  • twelve
  • seventeen
  • eighteen

If we replace ‘three’ and ‘five’ by the number of letters in these words, we obtain the equation:

5+4=x.

Hence x=9. So we need to replace x by a word with 9 letters in it. Of the options we are given ‘seventeen’ is the only word with 9 letters in it.

6.

5 marks

6.1 The square ABCD has an area of 196.

It contains two overlapping squares; the larger of these squares has an area 4 times that of the smaller and the area of their overlap is 1.

What is the total area of the shaded region?

  • 44
  • 72
  • 80
  • 152
  • more information is needed.
Show Hint (–1 mark)
–1 mark

We let P,Q,R,S,T and U be the points as labelled in the diagram. Since the square ABCD has an area of 196, it has side length 14. We suppose that the smaller square inside ABCD has side length x. Since the larger square inside ABCD has area 4 times that of the smaller square it will have side length 2x.

We let P,Q,R,S,T and U be the points as labelled in the diagram. Since the square ABCD area 196, it has side length 14. We suppose that the smaller square inside ABCD has side length x. Since the larger square inside ABCD has area 4 times that of the smaller square it will have side length 2x.

We note that the diagram is symmetric about the line AC. Hence the overlap of the squares is itself a square. Since this square has area 1, then it has side length 1. So QR has length 1. As UR=AP, UR has length x. Now QS has length 2x. Therefore US has length x+2x1=3x1. The length of US is the same as the side length of the square ABCD. Therefore 3x1=14. Hence x=5.

So the larger square inside ABCD has side length 10 and hence area 100, and the smaller square has area 25. As their overlap has area 1, the total area they cover is 100 + 25 − 1 = 124. Therefore, the shaded area is 196 − 124 = 72.

7.

5 marks

7.1 The diagrams show squares placed inside two identical semicircles. In the lower diagram the two squares are identical.

What is the ratio of the areas of the two shaded regions?

  • 1 : 2
  • 2 : 3
  • 3 : 4
  • 4 : 5
  • 5 : 6
Show Hint (–1 mark)
–1 mark

Choose units so that the two semicircles have radius 1. We let x be the side length of the square in the first diagram, and y be the side lengths of the squares in the second diagram.

We choose units so that the two semicircles have radius 1. We let x be the side length of the square in the first diagram, and y be the side lengths of the squares in the second diagram.

By Pythagoras’ Theorem, we have x2+12x2=1 and y2+y2=1. That is 54x2=1 and 2y2=1. Therefore x2=45 and y2=12.

The shaded region in the top diagram consists of a square of side length x and hence of area x2, that is, 45. The shaded region in the lower diagram consists of two squares each of side length y, and hence it has area 2y2, that is, 1. Hence the ratio of the two shaded regions is 45:1=4:5.

8.

4 marks

8.1 Four brothers are discussing the order in which they were born. Two are lying and two are telling the truth. Which two are telling the truth?

Alfred: “Bernard is the youngest.”
Horatio: “Bernard is the oldest and I am the youngest.”
Inigo: “I was born last.”
Bernard: “I’m neither the youngest nor the oldest.”

  • Bernard and Inigo
  • Horatio and Bernard
  • Alfred and Horatio
  • Alfred and Bernard
  • Inigo and Horatio
Show Hint (–1 mark)
–1 mark

Take a look at what Horatio says – does it fit with what any of the others say?

What Horatio says contradicts each of the statements made by his brothers. So he cannot be one of the two who are telling the truth.

What Alfred says contradicts what both what Inigo says and what Bernard says. So the two who are telling the truth cannot include Alfred.

We deduce that the two brothers who are telling the truth are Bernard and Inigo. This is indeed possible. For example, if Alfred was born first, Horatio second, Bernard third and Inigo last, then just Bernard and Inigo are telling the truth.

9.

5 marks

9.1 The diagram shows a shaded shape bounded by circular arcs with the same radius. The centres of three arcs are the vertices of an equilateral triangle; the other three centres are the midpoints of the sides of the triangle.

The sides of the triangle have length 2.

What is the difference between the area of the shaded shape and the area of the triangle?

  • π6
  • π4
  • π3
  • π2
  • π
Show Hint (–1 mark)
–1 mark

One way to think about this problem is that the difference between the area of the shaded shape and the area of the triangle is:

  • Shaded areas – Triangle area
  • (Shaded areas outside Triangle + shaded areas inside Triangle) – Triangle area
  • (Shaded areas outside Triangle + [Triangle area – 3 semicircles]) – Triangle area
  • Shaded areas outside Triangle + Triangle area – 3 semicircles – Triangle area
  • Shaded areas outside Triangle – 3 semicircles

One way to think about this problem is that the difference between the area of the shaded shape and the area of the triangle is:

  • Shaded areas – Triangle area
  • (Shaded areas outside Triangle + shaded areas inside Triangle) – Triangle area
  • (Shaded areas outside Triangle + [Triangle area – 3 semicircles]) – Triangle area
  • Shaded areas outside Triangle + Triangle area – 3 semicircles – Triangle area
  • Shaded areas outside Triangle – 3 semicircles

Since the sides of the triangle have length 2, each of the circular arcs has radius 12. The difference between the area of the shaded shape and that of the triangle, is the difference between the area of the 3 shaded sectors of circles outside the triangle, and that of the 3 unshaded semicircles within the triangle.

The angles of the equilateral triangle are 60° which is 16th of a complete revolution. So the areas of the shaded sectors of circles outside the triangle are 56ths of the total areas of these circles, each of which has radius 12.

So the area of these sectors is 3×56π122=58π. The area of the 3 unshaded semicircles inside the triangle is 3×12π122=38π. The difference between these is 58π38π=14π. Hence, the is the difference between the area of the shaded shape and the area of the triangle is π4.

10. 580-ton monster machine is building bridges across China

This machine is incredible. As you watch the video, look at when it moves the massive bridge sections forwards... what does it have to do before it can redistribute the weight?

Before you hit the SUBMIT button, here are some quick reminders:

  • You will receive your score immediately, and collect your reward points.
  • You might earn a new badge... if not, then maybe next week.
  • Make sure you go through the solution sheet – it is massively important.
  • A score of less than 50% is ok – it means you can learn lots from your mistakes.
  • The next Parallelogram is after the Easter holidays.
  • Finally, if you missed any earlier Parallelograms, make sure you go back and complete them. You can still earn reward points and badges by completing missed Parallelograms.

Cheerio, Simon.