| No Solution Yet | Submitted by Danish Ahmed Khan |
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Solution with assumptions
|
 | Comment 1 of 2 |
Two things I assume but didn't prove:
1) P and Q should be on the londer sides of the triangle.
2) PQ is minimized if PC=QC.
Let P be on AC and Q on BC. Call PC=QC=x
The area of ABC = 0.5ab*sin(C)
The area of PQC = 0.5xx*sin(C)
From which ab=2x^2
From the law of cosines for angle C
cos(C)=(a^2+b^2-c^2)/(2ab)
Using the law of cosines for PQ:
PQ^2 = x^2 + x^2 - 2xx*Cos(C)
Substitute to get
PQ^2 = ab - ab*Cos(C)
PQ^2 = ab - (a^2+b^2-c^2)/2
this can be rewritten as
PQ^2 = (c^2-(a-b)^2)/2
| Posted by Jer on 2020-10-20 08:10:51 |
