You may also like

problem icon

At a Glance

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

problem icon

Contact

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?

problem icon

Gift of Gems

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.)

Areas and Ratios

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3


Preveina, from Crest Girls' Academy, made this observation:

When triangles have their bases on the same line and the shared side (making the height the same) the area of the triangles and the length of the bases of their triangles are in the same ratio.



Mark, from Gozo college, explained his thinking for the two questions and the Final Challenge:



Question A

As regards triangles with their bases on the same line and a shared side, one observes that:

The ratio of the triangles' bases is equal to the ratio of their areas.

To prove this: Both triangles have the same perpendicular height. Let this be $h$. Let the area of one triangle be $A_1$ and the other $A_2$. Therefore, the base of one triangle is $\frac{2A_1}{h}$ and the other is $\frac{2A_2}{h}$.



So the ratio of base lengths is: $\frac{2A_1}{h} : \frac{2A_2}{h}$, which is equal to $A_1 : A_2$



Question B

Ratio of lengths red : blue is $1 : 2$

Therefore, ratio of areas is $1 : 2$, equivalent to $(A+B) : 8$. Therefore $A+B = \frac{8}{2} = 4$



For the $8 : 16$ triangles, $\frac{C+3}{D} = \frac{8}{16}$ so $C = \frac{D-6}{2}$

For the $8 : 10$ triangles, $\frac{C+3}{D} = \frac{8}{10}$ so $C = \frac{4D-15}{5}$



Final challenge: This builds on the principle found in question A and uses simultaneous equations. Let Area of triangle AOP = $x$, Let Area of triangle AOQ = $y$.



Treating triangle ABC as triangles ABQ and BCQ sharing a base: $\frac{x+8}{y} = \frac{BO}{OQ} = \frac{10}{5} = 2$ so $x = 2y-8$



Treating triangle ABC as triangles APC and BCP sharing a base: $\frac{x}{y+5} = \frac{OP}{CO} = \frac{8}{10} = \frac{4}{5}$ so $x = \frac{4y + 20}{5}$



Solving this pair of simultaneous equations gives

$2y-8 = \frac{4y+20}{5} \Rightarrow y = 10$

Substituting $y = 10$: $ x = 2\times10-8 \Rightarrow x = 12$

Therefore, area of quadrilateral $APOQ = x + y = 10 + 12 = 22$ square units.



Well done to Horizon who sent in a solution using similar ideas to Mark's.