, 2013

Since $\dfrac{21}{x}=\dfrac{7}{y}$, then $21 = \dfrac{7x}{y}$ or $\dfrac{x}{y}=\dfrac{21}{7}=3$. Solution 1 Since $$\dfrac{1}{3} \approx 0.3333 \qquad \dfrac{1}{4} = 0.25 \qquad \dfrac{1}{5} = 0.2 \qquad \dfrac{1}{6} \approx 0.1667$$ then $\dfrac{1}{5} < 0.2013$ and $0.2013 < \dfrac{1}{4}$, so $n$ must equal 4. (We should note as well that $\dfrac{1}{n}$ decreases as $n$ increases, so this is the only integer value of $n$ that works.) Solution 2 Since $\dfrac{1}{n+1}<0.2013$, then $n+1 > \dfrac{1}{0.2013}$ or $n > \dfrac{1}{0.2013}-1 \approx 3.9677$. Since $\dfrac{1}{n} > 0.2013$, then $n < \dfrac{1}{0.2013} \approx 4.9677$. Since $n$ is a positive integer that is smaller than a number that is approximately 4.9677, and that is larger than a number that is approximately 3.9677, then $n=4$. Since $AH$ is perpendicular to $BC$, then the area of $\triangle ABC$ equals $\frac{1}{2}(BC)(AH)$. Since we are told that this area equals 84 and $AH=8$, then $84 = \frac{1}{2}(BC)(8)$ or $4\cdot BC = 84$ or $BC=21$. Also, since $\triangle AHB$ is right-angled at $H$, then by the Pythagorean Theorem, $$BH = \sqrt{AB^2-AH^2}=\sqrt{10^2-8^2}=\sqrt{36}=6$$ since $BH>0$. (We could also have recognized two sides of a 6-8-10 right-angled triangle.) Since $BC=21$ and $BH=6$, then $HC = BC-BH=21-6=15$. Since $\triangle AHC$ is right-angled at $H$, then by the Pythagorean Theorem, $$AC = \sqrt{AH^2 + HC^2} = \sqrt{8^2+15^2}=\sqrt{289} = 17$$ since $AC>0$. Finally, the perimeter of $\triangle ABC$ equals $AB+BC+AC$ or $10+21+17$, which equals 48.