Difference between revisions of "1971 Canadian MO Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | + | Let the top of the flagpole of height h be denoted by H and its intersection with the surface be denoted by I. Similarly, let the top of the flagpole of length k be denoted by K and its intersection with the surface be denoted by J. Finally, let the desired point be P. Reflect H over IJ to H'. Since angles are preserved in reflections, we have < HPI = < H'PI. However, the problem statement tells us that < HPI = < KPJ. By transitivity, we must have < H'PI = < KPJ, implying that H', P, and K are collinear by the converse of Vertical Angles. In effect, we can construct our desired point P by reflecting H' over IJ and finding the intersection of H'K and IJ. Therefore, there is only one such point that is constructed with the method shown above. | |
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| + | -Solution by '''thecmd999''' | ||
== See Also == | == See Also == | ||
Latest revision as of 17:09, 19 September 2012
Problem
Two flag poles of height 
and 
are situated 
units apart on a level surface. Find the set of all points on the surface which are so situated that the angles of elevation of the tops of the poles are equal.
Solution
Let the top of the flagpole of height h be denoted by H and its intersection with the surface be denoted by I. Similarly, let the top of the flagpole of length k be denoted by K and its intersection with the surface be denoted by J. Finally, let the desired point be P. Reflect H over IJ to H'. Since angles are preserved in reflections, we have < HPI = < H'PI. However, the problem statement tells us that < HPI = < KPJ. By transitivity, we must have < H'PI = < KPJ, implying that H', P, and K are collinear by the converse of Vertical Angles. In effect, we can construct our desired point P by reflecting H' over IJ and finding the intersection of H'K and IJ. Therefore, there is only one such point that is constructed with the method shown above.
-Solution by thecmd999
See Also
| 1971 Canadian MO (Problems) | ||
| Preceded by Problem 8 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 10 |
