, 2014

The three angles shown in the pie chart are $(2x)^{\circ}$, $(3x)^{\circ}$ and $90^{\circ}$.

Since these three angles form a complete circle, then $(2x)^{\circ}+(3x)^{\circ}+90^{\circ}=360^{\circ}$, or $5x=270$ and so $x=54$.
The ratio of the number of bronze medals to the number of silver medals to the number of gold medals is equal to the ratio of the sector angles, $(2x)^{\circ}$ to $(3x)^{\circ}$ to $90^{\circ}$, respectively.

Since $x=54$, then the required ratio is $2(54):3(54):90$ or $108:162:90$.

Dividing each term by 18 the ratio becomes $6:9:5$, which is written in lowest terms.
Since the ratio of the number of bronze to silver to gold medals is $6:9:5$, let the number of bronze, silver and gold medals in the trophy case be $6x$, $9x$ and $5x$ respectively.

Since the total number of medals in the trophy case is 80, then $6x+9x+5x=80$ or $20x=80$ and so $x=4$.

Thus, there are $6\times4=24$ bronze medals, $9\times4=36$ silver medals, and $5\times4=20$ gold medals in the trophy case.
The trophy case begins with 24, 36 and 20 bronze, silver and gold medals, respectively.

Recall that the number of medals is in the ratio $6:9:5$.

For the ratio of the final number of medals to remain unchanged, we claim that the number of medals added by the teacher must also be in the ratio $6:9:5$.

(We will prove this claim is true at the end of the solution.)

Since $6:9:5$ is in lowest terms, the smallest number of medals that the teacher could have added is 6 bronze, 9 silver and 5 gold.

Therefore, the smallest number of medals that could now be in the trophy case is $80+6+9+5$ or 100 medals.

We note that the number of bronze, silver and gold medals is now 30, 45 and 25, which is still in the ratio $6:9:5$.

Proof of Claim: Let the number of bronze, silver and gold medals added be $b,s$ and $g$ respectively. When these are added to the existing medals, the number of bronze, silver and gold medals becomes $(24+b)$, $(36+s)$ and $(20+g)$. The claim is that for the new ratio, $(24+b):(36+s):(20+g)$, to remain unchanged (that is, to equal $24:36:20$), then $b:s:g$ must equal $6:9:5$. If $(24+b):(36+s):(20+g)=24:36:20$, then

$\frac{36+s}{24+b}=\frac{36}{24}$ and $\frac{20+g}{36+s}=\frac{20}{36}$. From the first equation, $24(36)+24s=36(24)+36b$ and

so $24s=36b$ or $\frac{s}{b}=\frac{36}{24}=\frac96$. Similarly, from the second equation we can show that $\frac{g}{s}=\frac59$. Thus, $b:s:g=6:9:5$ as claimed.