https://artofproblemsolving.com/wiki/index.php?title=2002_Indonesia_MO_Problems/Problem_7&feed=atom&action=history2002 Indonesia MO Problems/Problem 7 - Revision history2024-03-29T05:41:54ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2002_Indonesia_MO_Problems/Problem_7&diff=132051&oldid=prevDuck master: created page with solution & categorization2020-08-18T16:41:16Z<p>created page with solution & categorization</p>
<p><b>New page</b></p><div>Let <math>ABCD</math> be a rhombus with <math>\angle A = 60^\circ</math>, and <math>P</math> is the intersection of diagonals <math>AC</math> and <math>BD</math>. Let <math>Q</math>, <math>R</math>, and <math>S</math> are three points on the rhombus' perimeter. If <math>PQRS</math> is also a rhombus, show that exactly one of <math>Q</math>, <math>R</math>, and <math>S</math> is located on the vertices of rhombus <math>ABCD</math>.<br />
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== Solution ==<br />
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Firstly, all rhombi are [[parallelogram|parallelograms]], so that <math>P</math> is the [[centroid]] of <math>ABCD</math>.<br />
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Suppose that <math>Q, R, S</math> are all on one side of the rhombus. Then, in order for <math>PQRS</math> to be a parallelogram, <math>P</math> should also be on that side. But this is not so, so this case is impossible.<br />
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Suppose that <math>Q, R, S</math> are on two sides of the rhombus; then one side is occupied by two of these points (the "majority side") and one side is occupied by only one of these points (the "minority side"). If <math>R</math> is on the minority side, then <math>PQRS</math> is necessarily self-intersecting and thusly not a parallelogram. Thusly, either <math>Q</math> or <math>S</math> is on the minority side; WLOG it is <math>Q</math>. Then <math>\vec{SR}</math> is parallel to the majority side, so <math>\vec{PQ}</math> must also be parallel to the majority side, so that <math>Q</math> is the midpoint of the minority side. Then <math>\vec{PQ} = \vec{SR}</math> must be exactly half the length of the majority side.<br />
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From here, we consider cases. Based on the symmetry of <math>ABCD</math>, however, we only need consider two: that where the majority side is <math>AB</math> and the minority side <math>BC</math>, and that where the majority side is <math>AB</math> and the minority side <math>DA</math>. In the first case, we find that in order to satisfy <math>PQ = QR</math> and <math>A, R, B</math> [[collinear]], we must have <math>R = B</math> or <math>R</math> be outside of the segment <math>AB</math>, which is forbidden, so that exactly one vertex of <math>PQRS</math> (<math>R</math>) is also a vertex of <math>ABCD</math>. In the second case, we find that in order to satisfy <math>PQ = QR</math> and <math>A, R, B</math> collinear, we must have <math>R = A</math> or <math>R</math> be the midpoint of <math>AB</math>, so that exactly one vertex of <math>PQRS</math> (<math>R</math> or <math>S</math>, respectively) is also a vertex of <math>ABCD</math>. <br />
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Finally, suppose that <math>Q, R, S</math> are on three different sides. WLOG, suppose that <math>R\in AB</math>. If one of the other vertices is on <math>CD</math> (WLOG it is <math>Q</math>), then <math>S</math> must be outside the parallelogram (since <math>h_S = h_P - h_Q + h_R = \frac{1}{2} - 1 + 0 = -\frac{1}{2}</math>, where <math>h_X</math> is the (signed) [[height]] of <math>X</math> to <math>AB</math>, scaled by the height of <math>C</math>). This is impossible, so we know that <math>Q</math> and <math>S</math> must not be on <math>CD</math>; WLOG, we have <math>Q\in DA, S\in BC</math>. Then the midpoint of <math>QS</math> is on the line halfway between lines <math>DA</math> and <math>BC</math>. Since the midpoint of <math>QS</math> and that of <math>PR</math> are the same, <math>R</math> is the midpoint of <math>AB</math>. Then, in order to satisfy <math>PQ = QR</math> and <math>PS = SR</math>, we must have <math>Q</math> the midpoint of <math>DA</math> and <math>S = B</math>, so that exactly one vertex of <math>PQRS</math> (that is, <math>S</math>) is also a vertex of <math>ABCD</math>.<br />
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All cases having been considered, we have shown that if <math>PQRS</math> is a rhombus, then exactly one of <math>Q</math>, <math>R</math>, and <math>S</math> is a vertex of <math>ABCD</math>, and we are done.<br />
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== See also ==<br />
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{{Indonesia MO box|year=2002|num-b=6|after=Last Problem}}<br />
[[Category:Geometry Problems]]</div>Duck master