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.
(I used v for square root because it's quicker)
Let S be the midpoint of QR
Let M be the intersection of PX and EF
Let N be the intersection of PX and PS
Let a be the measure of angle QPX
Let b be the measure of angle XPS
by pythagorean theorem
PS = 5v15
PX = 8v6
and since the fold find the midpoint of PX,
PM = 4v6
also PME is a right angle formed by the fold
By similar triangles PSX and PMN
MN = 12v10/25
and PN = 64v15/25
By triangle PSR, sin(a+b)=1/4
By triangle PSX, sin(b)= v6/16
from this derive sin(a)=v10/32
derive cos(a+2b) = 3v6/8
Triangle PME can now be solved. EM=4v10/13
Triangle PNF can now be solved. NF=64v10/75
EF = EM+MN+NF = 64v10/39
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Posted by Jer
on 2015-03-19 14:16:21 |
exact solution
