$ ABC$ and $ A'B'C'$ are two triangles in the same plane such that the lines $ AA',BB',CC'$ are mutually parallel. Let $ [ABC]$ denotes the area of triangle $ ABC$ with an appropriate $ \pm$ sign, etc.; prove that \[ 3([ABC] + [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B'] + [A'BC] + [B'CA] + [C'AB].\]
Problem
Source: 1977 USAMO Problem 2
Tags: geometry, vector
keyree10
04.04.2010 21:17
I suppose the $ \pm$ thing is clockwise and anticlockwise (?) A convenient way of dealing with the signs is through vectors. We represent the signed area $ [XYZ]$ as $ \frac 1 2 \vec{XY}\times\vec{YZ}$. Now, after multiplying throughout by two, the $ \text{RHS}$ becomes $ \sum_{cyc}\vec{AB'}\times\vec{B'C'} + \vec{A'B}\times\vec{BC} = \sum_{cyc}(\vec{AA'} + \vec{A'B'})\times\vec{B'C'} + (\vec{A'A} + \vec{AB})\times\vec{BC} = \text{LHS} + \sum_{cyc}\vec{AA'}\times(\vec{B'C'} - \vec{BC})$ Now, Using $ \vec{PQ} = \vec{Q} - \vec{P}$ (w.r.t. some arbitrary origin) we can expand the extra terms and show that they cancel out.
Any suggestions on how to make the arrow (denoting vector) big enough to cover the term underneath?
iarnab_kundu
26.04.2013 09:19
Sure, even though its too late. the code is \overrightarrow{XYZ} Output: $\overrightarrow{XYZ}$
huashiliao2020
15.04.2023 02:55
Are there any other synthetic solutions to this? I'm not sure how to start, nor do I understand what $\pm$ means. @below ok but like what I mean how can area be plus minus
cinnamon_e
15.04.2023 03:10
huashiliao2020 wrote: Are there any other synthetic solutions to this? I'm not sure how to start, nor do I understand what $\pm$ means. $\pm$ means plus or minus, so for example $\pm 4=4$ or $-4$