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Isosceles Crease (Posted on 2015-03-19) Difficulty: 3 of 5
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    
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Solution 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

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