Difference between revisions of "1982 IMO Problems/Problem 5"
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We can assign coordinates to solve this question. First, WLOG, let the side length of the hexagon be 1. We also know that each angle of the hexagon is <math>\frac{(6-2) \cdot 180}{6} = 120</math>. From Law of Cosines, we find <math>AC = \sqrt{1 + 1 - 2cos(120)} = \sqrt3</math>. | We can assign coordinates to solve this question. First, WLOG, let the side length of the hexagon be 1. We also know that each angle of the hexagon is <math>\frac{(6-2) \cdot 180}{6} = 120</math>. From Law of Cosines, we find <math>AC = \sqrt{1 + 1 - 2cos(120)} = \sqrt3</math>. | ||
− | Now, let point <math>E</math> be located at <math>(0, 0)</math>. Since <math>AE = AC, A</math> is located at <math>(0, \sqrt3)</math> meaning <math>B</math> is located at <math>(1, \sqrt3)</math>. Using the ratios given to us, <math>CN = r\sqrt3</math> so <math>EN = (1-r)\sqrt3</math>. Let <math>O</math> be defined on line <math>ED</math> such that <math>EO</math> is perpendicular to <math>NO</math>. Since <math>\bigtriangleup ENO</math> is a <math>30-60-90</math> triangle, we can find that point N is located at <math>(\frac{3(1-r)}{2}, \frac{(1-r)\sqrt3}{2})</math>. Similarly, <math>M</math> would be located at <math>(\frac{3r}{2}, \frac{\sqrt3(1 - r)}{2})</ | + | Now, let point <math>E</math> be located at <math>(0, 0)</math>. Since <math>AE = AC,</math> <math>A</math> is located at <math>(0, \sqrt3)</math> meaning <math>B</math> is located at <math>(1, \sqrt3)</math>. Using the ratios given to us, <math>CN = r\sqrt3</math> so <math>EN = (1-r)\sqrt3</math>. Let <math>O</math> be defined on line <math>ED</math> such that <math>EO</math> is perpendicular to <math>NO</math>. Since <math>\bigtriangleup ENO</math> is a <math>30-60-90</math> triangle, we can find that point N is located at <math>(\frac{3(1-r)}{2}, \frac{(1-r)\sqrt3}{2})</math>. Similarly, <math>M</math> would be located at <math>(\frac{3r}{2}, \sqrt3(1 - \frac{r}{2}))</math>. |
+ | |||
+ | Since B, M, N, are collinear, they have the same slope. So, | ||
+ | <cmath>\frac{\sqrt3 - \frac{(1-r)\sqrt3}{2}}{1 - \frac{3(1-r)}{2}} = \frac{\sqrt3 - \sqrt3(1 - \frac{r}{2})}{1 - \frac{3r}{2}}</cmath> | ||
+ | <cmath>\frac{\sqrt3 + r\sqrt3}{3r - 1} = \frac{r\sqrt3}{2 - 3r}</cmath> | ||
+ | <cmath>2\sqrt3 + 2r\sqrt3 - 3r\sqrt3 - 3r^2\sqrt3 = 3r^2\sqrt3 - r\sqrt3</cmath> | ||
+ | <cmath>6r^2\sqrt3 = 2\sqrt3</cmath> | ||
+ | <cmath>\boxed{r = \frac{\sqrt3}{3}}</cmath> | ||
+ | ~SID-DARTH-VATER |
Latest revision as of 22:05, 1 December 2024
Problem
The diagonals and of the regular hexagon are divided by inner points and respectively, so thatDetermine if and are collinear.
Solution 1
O is the center of the regular hexagon. Then we clearly have . And therefore we have also obviously , as . So we have and . Because of the quadrilateral is cyclic. . And as we also have we get . . And as we get .
This solution was posted and copyrighted by Kathi. The original thread for this problem can be found here: [1]
Solution 2
Let be the intersection of and . is the mid-point of . Since , , and are collinear, then by Menelaus Theorem, . Let the sidelength of the hexagon be . Then . Substituting them into the first equation yields
This solution was posted and copyrighted by leepakhin. The original thread for this problem can be found here: [2]
Solution 3
Note . From the relation results , i.e.
. Thus,
Therefore, , i.e.
This solution was posted and copyrighted by Virgil. The original thread for this problem can be found hercommunity/p398343]
Solution 4
Let . By the cosine rule,
.
.
Now if B, M, and N are collinear, then
.
By the law of Sines,
.
Also,
.
But
, which means . So, r = .
Solution 5
We can assign coordinates to solve this question. First, WLOG, let the side length of the hexagon be 1. We also know that each angle of the hexagon is . From Law of Cosines, we find .
Now, let point be located at . Since is located at meaning is located at . Using the ratios given to us, so . Let be defined on line such that is perpendicular to . Since is a triangle, we can find that point N is located at . Similarly, would be located at .
Since B, M, N, are collinear, they have the same slope. So, ~SID-DARTH-VATER