Geometricum geometriorum (HP IV, Goblet of fire)

Geometry Level 3

The right- triangle Δ A B C \Delta ABC shown above, ( A = 90 º \angle A = 90º ), has c = A B = 15 c = AB = 15 , and C D = m = 16 CD = m = 16 . Find the altitude (height) h h from A A to B C BC with foot at D D , the lenght of side b = A C b = AC and n = B D n = BD .

Submit b + h + n b + h + n


The answer is 41.

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1 solution

Viki Zeta
Oct 13, 2016

A B C = x A C B = 90 x In Δ A B D B A D = 90 x = A C B In Δ A C D C A D = x = A B C In Δ A B C and Δ D B A B A C = A D B = 90 B A D = A C B Δ A B C Δ D B A (AA) A B B D = B C A B A B 2 = B C A B 1 5 2 = n ( n + 16 ) 225 = n ( n + 16 ) n = 9 , 25 n = 9 In Δ A B C B C 2 = A B 2 + A C 2 b = A C 2 = B C 2 A B 2 = 400 b = 20 In Δ A D B Δ A B C Δ D B A A B D B = A C A D 15 9 = 20 h h = 12 n + b + h = 12 + 20 + 9 = 41 \angle ABC = x \\ \implies \angle ACB = 90-x \\ \text{In } \Delta ABD \\ \boxed{\angle BAD = 90 - x = \angle ACB} \\ \text{In } \Delta ACD \\ \boxed{\angle CAD = x = \angle ABC} \\ \text{In } \Delta ABC \text{ and } \Delta DBA \\ \angle BAC = \angle ADB = 90 \\ \angle BAD = \angle ACB \\ \implies \Delta ABC ~ \Delta DBA \text{ (AA)} \\ \implies \dfrac{AB}{BD} = \dfrac{BC}{AB} \\ \implies AB^2 = BC \cdot AB \\ \implies 15^2 = n(n+16) \\ \implies 225 = n(n+16) \\ \implies n = 9, -25 \\ \implies \color{#20A900}{\boxed{n = 9}} \\ \text{In }\Delta ABC \\ BC^2 = AB^2 + AC^2\\ \implies b = AC^2 = BC^2 - AB^2 = 400 \\ \implies \color{#D61F06}{\boxed{b = 20}} \\ \text{In } \Delta ADB \\ \Delta ABC ~ \Delta DBA \\ \implies \dfrac{AB}{DB} = \dfrac{AC}{AD} \\ \implies \dfrac{15}{9} = \dfrac{20}{h} \\ \implies \color{#3D99F6}{\boxed{h = 12}} \\ \color{magenta}{\boxed{\therefore n + b + h = 12 + 20 + 9 = 41}}

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