The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?

A circular plate rolls in contact with the sides of a rectangular tray. How much of its circumference comes into contact with the sides of the tray when it rolls around one circuit?

Four jewellers possessing respectively eight rubies, ten saphires, a hundred pearls and five diamonds, presented, each from his own stock, one apiece to the rest in token of regard; and they thus became owners of stock of precisely equal value. Tell me the prices of their gems and the values of their stocks. (Editors note: In this problem you are only given enough information to find the values of rubies, saphires and diamonds relative to the price of a pearl.)

This problem is in two parts. The first part provides two building blocks which will help you to solve the final challenge. These can be attempted in either order. Of course, you are welcome to go straight to the Final Challenge!

Click on a question below to get started.

Question A

This question is about triangles with their bases on the same line and a shared side, like the ones below:

The numbers in the triangles represent their areas.

The diagrams are not drawn to scale - can you create diagrams with the correct areas drawn to scale?

Convince yourself that there are many different possibilities

Make a note of the base lengths of your triangles.

Notice anything interesting? Convince yourself it always happens.

The numbers in the triangles represent their areas.

The diagrams are not drawn to scale - can you create diagrams with the correct areas drawn to scale?

Convince yourself that there are many different possibilities

Make a note of the base lengths of your triangles.

Notice anything interesting? Convince yourself it always happens.

Question B

The red line is half the length of the blue line. Numbers and letters inside a triangle represent its area. What can you say about $A+B$?

For each triangle with base lengths as shown, write C in terms of D.

FINAL CHALLENGE

Look at the diagram below (which is not drawn to scale).

*Click below to reveal a method for finding the area. The statements have been muddled up, however. Can you rearrange them into the correct order?*

A set of printable cards is also avilable.

A. Substituting this back into equation (1) gives $E = 12$.

B. Triangles $QAO$ and $OAB$ have the same height above the line $BQ$, so $\frac{F}{OQ}=\frac{E+8}{BO}$.

C. Triangles $PAO$ and $OAC$ have the same height above the line $PC$, so $\frac{E}{PO}=\frac{F+5}{CO}$.

D. Combining these two equations gives $\frac{E}{8} = \frac{F+5}{10}$.

E. The area of quadrilateral $AQOP$ is $E+F = 10+12=22$.

F. Triangles $PBO$ and $OBC$ also have the same height above $PC$, so $\frac{8}{PO}=\frac{10}{CO}$.

G. Solving for $F$ gives $F=10$.

H. Doubling equation (2) and rearranging gives $10E = 20F - 80$.

I. Draw in the line $AO$. Label the area of triangle $AOP$ as $E$, and that of $AOQ$ as $F$.

J. Clearing denominators gives $10E = 8F+40$. Call this equation (1).

K. Clearing denominators gives $10F = 5E+40$. Call this equation (2).

L. Combining these two equations gives $\frac{F}{5} = \frac{E+8}{10}$.

M. Triangles $QCO$ and $OBC$ also have the same height above $BQ$, so $\frac{5}{OQ}=\frac{10}{BO}$.

N. Substituting this into equation (1) gives $20F - 80 = 8F+40$.

The areas of three of the triangles are shown.

What is the area of the quadrilateral $APOQ$?

A set of printable cards is also avilable.

A. Substituting this back into equation (1) gives $E = 12$.

B. Triangles $QAO$ and $OAB$ have the same height above the line $BQ$, so $\frac{F}{OQ}=\frac{E+8}{BO}$.

C. Triangles $PAO$ and $OAC$ have the same height above the line $PC$, so $\frac{E}{PO}=\frac{F+5}{CO}$.

D. Combining these two equations gives $\frac{E}{8} = \frac{F+5}{10}$.

E. The area of quadrilateral $AQOP$ is $E+F = 10+12=22$.

F. Triangles $PBO$ and $OBC$ also have the same height above $PC$, so $\frac{8}{PO}=\frac{10}{CO}$.

G. Solving for $F$ gives $F=10$.

H. Doubling equation (2) and rearranging gives $10E = 20F - 80$.

I. Draw in the line $AO$. Label the area of triangle $AOP$ as $E$, and that of $AOQ$ as $F$.

J. Clearing denominators gives $10E = 8F+40$. Call this equation (1).

K. Clearing denominators gives $10F = 5E+40$. Call this equation (2).

L. Combining these two equations gives $\frac{F}{5} = \frac{E+8}{10}$.

M. Triangles $QCO$ and $OBC$ also have the same height above $BQ$, so $\frac{5}{OQ}=\frac{10}{BO}$.

N. Substituting this into equation (1) gives $20F - 80 = 8F+40$.

The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the
NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to
embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities
can be found here.

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NRICH is part of the family of activities in the Millennium Mathematics Project.

NRICH is part of the family of activities in the Millennium Mathematics Project.