PRMO 2017 question 17
17. Suppose the altitudes of a triangle are 10, 12 and 15. What is its semi-perimeter?
Solution:
First, denote the area of the triangle by Δ, and let its sides be \(a,b,c\) opposite the altitudes \(h_a=10,\;h_b=12,\;h_c=15\). Then by definition of altitude, \[ a=\frac{2Δ}{h_a},\quad b=\frac{2Δ}{h_b},\quad c=\frac{2Δ}{h_c}. \] Hence the semiperimeter is \[ s=\frac{a+b+c}{2} =\frac1{2}\Bigl(\frac{2Δ}{10}+\frac{2Δ}{12}+\frac{2Δ}{15}\Bigr) =Δ\Bigl(\tfrac1{10}+\tfrac1{12}+\tfrac1{15}\Bigr) =Δ\cdot\frac{1}{4} =\frac{Δ}{4}. \] On the other hand, by Heron’s formula, \[ Δ^2 =s(s-a)(s-b)(s-c), \] and one checks that \[ s-a=\frac{Δ}{4}-\frac{2Δ}{10}=\frac{Δ}{20},\quad s-b=\frac{Δ}{12},\quad s-c=\frac{7Δ}{60}. \] Plugging in gives \[ Δ^2 =\frac{Δ}{4}\;\frac{Δ}{20}\;\frac{Δ}{12}\;\frac{7Δ}{60} =\frac{7\,Δ^4}{57600} \;\;\Longrightarrow\;\; Δ^2=\frac{57600}{7} \;\;\Longrightarrow\;\; Δ=\frac{240}{\sqrt7}. \] Therefore \[ s=\frac{Δ}{4}=\frac{240}{4\sqrt7} =\frac{60}{\sqrt7} =\frac{60\sqrt7}{7} \approx22.68. \] **Answer.** The semiperimeter is \[ \boxed{\frac{60}{\sqrt7}\approx22.68}. \]
First, denote the area of the triangle by Δ, and let its sides be \(a,b,c\) opposite the altitudes \(h_a=10,\;h_b=12,\;h_c=15\). Then by definition of altitude, \[ a=\frac{2Δ}{h_a},\quad b=\frac{2Δ}{h_b},\quad c=\frac{2Δ}{h_c}. \] Hence the semiperimeter is \[ s=\frac{a+b+c}{2} =\frac1{2}\Bigl(\frac{2Δ}{10}+\frac{2Δ}{12}+\frac{2Δ}{15}\Bigr) =Δ\Bigl(\tfrac1{10}+\tfrac1{12}+\tfrac1{15}\Bigr) =Δ\cdot\frac{1}{4} =\frac{Δ}{4}. \] On the other hand, by Heron’s formula, \[ Δ^2 =s(s-a)(s-b)(s-c), \] and one checks that \[ s-a=\frac{Δ}{4}-\frac{2Δ}{10}=\frac{Δ}{20},\quad s-b=\frac{Δ}{12},\quad s-c=\frac{7Δ}{60}. \] Plugging in gives \[ Δ^2 =\frac{Δ}{4}\;\frac{Δ}{20}\;\frac{Δ}{12}\;\frac{7Δ}{60} =\frac{7\,Δ^4}{57600} \;\;\Longrightarrow\;\; Δ^2=\frac{57600}{7} \;\;\Longrightarrow\;\; Δ=\frac{240}{\sqrt7}. \] Therefore \[ s=\frac{Δ}{4}=\frac{240}{4\sqrt7} =\frac{60}{\sqrt7} =\frac{60\sqrt7}{7} \approx22.68. \] **Answer.** The semiperimeter is \[ \boxed{\frac{60}{\sqrt7}\approx22.68}. \]
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