MathDB

The dissection of the

3-4-5

triangle shown below has diameter

5/2

. [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(23cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 1.42, xmax = 24.42, ymin = 3.8, ymax = 15.54; /* image dimensions */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax,defaultpen+black, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true); yaxis(ymin, ymax,defaultpen+black, Ticks(laxis, Step = 1, Size = 2, NoZero), Arrows(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((9.44,8.52)--(12.44,8.52)); draw((9.44,12.52)--(9.44,8.52)); draw((9.44,12.52)--(12.44,8.52)); draw((9.44,10.52)--(10.94,10.52)); draw((10.94,10.52)--(10.94,8.52)); draw((9.44,8.52)--(10.94,10.52)); /* dots and labels */ dot((9.44,8.52),dotstyle); label("

A

", (9.52,8.64), NE * labelscalefactor); dot((12.44,8.52),dotstyle); label("

B

", (12.52,8.64), NE * labelscalefactor); dot((9.44,12.52),dotstyle); label("

C

", (9.52,12.64), NE * labelscalefactor); dot((10.94,8.52),dotstyle); label("

D

", (11.02,8.64), NE * labelscalefactor); dot((9.44,10.52),dotstyle); label("

E

", (9.52,10.64), NE * labelscalefactor); dot((10.94,10.52),dotstyle); label("

F

", (11.02,10.64), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy] Find the least diameter of this triangle into four parts. (The diameter of a dissection is the least upper bound of the distances between pairs of points belonging to the same part.)

Problem Statement