 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Isosceles Crease (Posted on 2015-03-19) Consider an isosceles triangle PQR with PQ = PR=20 and QR=10 . The vertex P is folded onto the point X in QR, forming the crease EF with E on PQ and F on PR.

Given that QX = 2, find the length of EF.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solution | Comment 1 of 2
Geometers' Sketchpad gives the length as 5.18938.

The second measurement at top left gives the conversion factor between cm and dimensionless units in the link here to the diagrm.

Let Y be the midpoint of QR.

angle QPY = asin(1/4)
|PY| = 20 cos(QPY)
angle XPY = atan(3/|PY|)
angle QPX = angle QPY - angle XPY
|PX| = 3/sin(XPY)
|PE| = 10/cos(QPX)
angle RPY = angle QPY
|EF| = (PX/2) tan(QPX) + (PX/2) tan(XPY+RPY)

20   QPY=asin(1/4)
30   PY=20*cos(QPY)
40   XPY=atan(3/PY)
50   QPX=QPY-XPY
60   PX=3/sin(XPY)
70   PE=10/cos(QPX)
80   RPY=QPY
90   EF=(PX/2)*tan(QPX)+(PX/2)*tan(XPY+RPY)
100   print EF

finds

5.1893787243788789014

Edited on March 19, 2015, 11:57 am
 Posted by Charlie on 2015-03-19 11:28:48 Please log in:

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