Six sets of all points P inside triangle ABC

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Six sets of all points P inside triangle ABC

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Source: Canadian Repêchage 2012: Problem 4

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TheMaskedMagician

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TheMaskedMagician

#1 h May 19, 2014, 9:21 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle such that $\angle BAC = 90^\circ$ and $AB < AC$ . We divide the interior of the triangle into the following six regions: $\begin{align*} S_1=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PA<PB<PC \\ S_2=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PA<PC<PB \\ S_3=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PB<PA<PC \\ S_4=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PB<PC<PA \\ S_5=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PC<PA<PB \\ S_6=\text{set of all points }\mathit{P}\text{ inside }\triangle ABC\text{ such that }PC<PB<PA\end{align*}$ Suppose that the ratio of the area of the largest region to the area of the smallest non-empty region is $49 : 1$ . Determine the ratio $AC : AB$ .

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#2 h May 23, 2014, 10:19 AM • 2 Y

Y by Adventure10, Mango247

Let $L,M,N$ be midpoints of $BC,CA,AB$ , perpendicular bisector of $BC$ intersects $AC$ at $D$ . Then $LN,LD,LM$ divide the triangle into 4 regions, which are the non-empty regions of all the $S_i$ . Then $[ADLN]:[DML]=49:1$ , which implies $AM=25MD$ . Then $AC/AB=5$ follows.

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