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.
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
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Posted by Charlie
on 2015-03-19 11:28:48 |
solution
