Difference between revisions of "1978 USAMO Problems/Problem 2"
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The point is obviously unique, because the two maps have different scales (but if P and Q where two fixed points the distance between them would be the same on both maps). | The point is obviously unique, because the two maps have different scales (but if P and Q where two fixed points the distance between them would be the same on both maps). | ||
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==Solution 2== | ==Solution 2== | ||
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The point is obviously unique because of the different scale and angle, which can also be seen as the point can and must be the unique center of spiral similarity taking ABCD to A'B'C'D', as under the spiral similarity the center does not move. (Recall that a spiral similarity is a dilation combined with a rotation, and that there is a unique spiral similarity taking any segment to another segment, provided the four points do not make a parallelogram.) | The point is obviously unique because of the different scale and angle, which can also be seen as the point can and must be the unique center of spiral similarity taking ABCD to A'B'C'D', as under the spiral similarity the center does not move. (Recall that a spiral similarity is a dilation combined with a rotation, and that there is a unique spiral similarity taking any segment to another segment, provided the four points do not make a parallelogram.) | ||
To construct this center of spiral similarity, reference EGMO Lemma 10.1. Draw line DD' and AA', and let them intersect at point X. Draw circles through points X,A,D and X,A',D' (using the perpendicular bisector constructions to find the center of the circle using two chords). The intersection point of the two circles other than X is the center of spiral similarity, and hence is O, which represents the same point on both maps. | To construct this center of spiral similarity, reference EGMO Lemma 10.1. Draw line DD' and AA', and let them intersect at point X. Draw circles through points X,A,D and X,A',D' (using the perpendicular bisector constructions to find the center of the circle using two chords). The intersection point of the two circles other than X is the center of spiral similarity, and hence is O, which represents the same point on both maps. | ||
Latest revision as of 20:57, 12 November 2023
Contents
Problem
and 
are square maps of the same region, drawn to different scales and superimposed as shown in the figure. Prove that there is only one point 
on the small map that lies directly over point 
of the large map such that and each represent the same place of the country. Also, give a Euclidean construction (straight edge and compass) for .
![[asy] defaultpen(linewidth(0.7)+fontsize(10)); real theta = -100, r = 0.3; pair D2 = (0.3,0.76); string[] lbl = {'A', 'B', 'C', 'D'}; draw(unitsquare); draw(shift(D2)*rotate(theta)*scale(r)*unitsquare); for(int i = 0; i < lbl.length; ++i) { pair Q = dir(135-90*i), P = (.5,.5)+Q/2^.5; label("$"+lbl[i]+"'$", P, Q); label("$"+lbl[i]+"$",D2+rotate(theta)*(r*P), rotate(theta)*Q); }[/asy]](http://latex.artofproblemsolving.com/6/f/a/6fa08871047fa32076357c574b19dfeb124210b6.png)
Solution 1
The point is obviously unique, because the two maps have different scales (but if P and Q where two fixed points the distance between them would be the same on both maps).
Let the small map square be A'B'C'D' and the large be ABCD, where X and X' are corresponding points. We deal first with the special case where A'B' is parallel to AB. In this case let AA' and BB' meet at O. Then triangles OAB and OA'B' are similar, so O must represent the same point. So assume A'B' is not parallel to AB.
Let the lines A'B' and AB meet at W, the lines B'C' and BC meet at X, the lines C'D' and CD meet at Y, and the lines D'A' and DA meet at Z. We claim that the segments WY and XZ meet at a point O inside the smaller square. W cannot lie between A' and B' (or one of the vertices A', B' of the smaller square would lie outside the larger square). If it lies on the opposite side of A' to B', then Y must lie on the opposite side of C' to D'. Thus the segment WY must cut the side A'D' at some point Z' and the side B'C' at some point X'. The same conclusion holds if W lies on the opposite side of B' to A', because then Y must lie on the opposite side of D' to C'. Similarly, the segment XZ must cut the side A'B' at some point W' and the side C'D' at some point Y'. But now the segments X'Z' and W'Y' join pairs of points on opposite sides of the small square and so they must meet at some point O inside the small square.
Now the triangles WOW' and YOY' are similar (WW' and YY' are parallel). Hence OW/OY = OW'/OY'. So if we set up coordinate systems with AB as the x-axis and AD as the y-axis (for the large square) and A'B' as the x'-axis and A'D' as the y'-axis (for the small square) so that corresponding points have the same coordinates, then the y coordinate of O equals the y' coordinate of O. Similarly, XOX' and ZOZ' are similar, so OX/OZ = OX'/OZ', so the x-coordinate of O equals its x'-coordinate. In other words, O represents the same point on both maps.
Solution 2
The point is obviously unique because of the different scale and angle, which can also be seen as the point can and must be the unique center of spiral similarity taking ABCD to A'B'C'D', as under the spiral similarity the center does not move. (Recall that a spiral similarity is a dilation combined with a rotation, and that there is a unique spiral similarity taking any segment to another segment, provided the four points do not make a parallelogram.) To construct this center of spiral similarity, reference EGMO Lemma 10.1. Draw line DD' and AA', and let them intersect at point X. Draw circles through points X,A,D and X,A',D' (using the perpendicular bisector constructions to find the center of the circle using two chords). The intersection point of the two circles other than X is the center of spiral similarity, and hence is O, which represents the same point on both maps.
See Also
| 1978 USAMO (Problems • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
