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Ken's Puzzle of the Week

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A Set of Triangles

In
rectangle ABCD, point E is on AB and point F is on AD. Angle
CEF is a right angle. Draw triangle CEF.
The rectangle is divided into four triangles (AEF, BEC, DCF,
CEF). All
three sides of each triangle are integers, and CF is the longest line
in the
diagram. What is the smallest possible length of CF?

Extension: What solutions
exist for CF < 100? In particular what solutions exist which
aren't multiples of smaller solutions?

Source: Original.

Solutions were received from Philippe Fondanaiche, David Madfes,
Joseph DeVincentis,
Bernie R. Erickson, Alex Ang, Claudio Baiocchi, Alex Doskey, Anthony
Nguyen.
Alex Doskey found 11 solutions meeting the requirements (12 if we allow
CF=100).
He also found that the first solution which doesn't use a 3-4-5 based
triangle has CF=221.
(AE,AF,EF=H1),(BE,BC,CE=H2),(DF,CD,CF=H3),(H1,H2,H3)

1 = (12,9,15),(12,16,20), (7,24,25),(15,20,25)

2 = (24,18,30),(24,32,40),(14,48,50),(30,40,50) = #1x2

3 = (15,20,25),(48,36,60),(16,63,65),(25,60,65)

4 = (20,15,25),(36,48,60),(33,56,65),(25,60,65)

5 = (48,20,52),(15,36,39),(16,63,65),(52,39,65)

6 = (36,15,39),(20,48,52),(33,56,65),(39,52,65)

7 = (36,27,45),(36,48,60),(21,72,75),(45,60,75) = #1x3

8 = (32,24,40),(45,60,75),(36,77,85),(40,75,85)

9 = (45,24,51),(32,60,68),(36,77,85),(51,68,85)

10= (24,32,40),(60,45,75),(13,84,85),(40,75,85)

11= (60,32,68),(24,45,51),(13,84,85),(68,51,85)

12= (48,36,60),(48,64,80),(28, 96,100),(80,80,100) = #1x4

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