Difference between revisions of "1967 IMO Problems/Problem 4"
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equivalently, the one with maximum side <math>BC</math>. So we will try to maximize <math>BC</math>. | equivalently, the one with maximum side <math>BC</math>. So we will try to maximize <math>BC</math>. | ||
− | [[File:Prob_1967_4_fig1.png| | + | The plan is to find the value of <math>\alpha</math> which maximizes <math>BC</math>. |
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+ | [[File:Prob_1967_4_fig1.png|600px]] | ||
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+ | Note that for any <math>\alpha</math> we can construct the line through <math>A_0</math> which | ||
+ | forms the angle <math>\alpha</math> with <math>A_0C_0</math>. We can construct points <math>B, C</math> | ||
+ | on this line, and lines through these points which form angles | ||
+ | <math>\angle B, \angle C</math> with the line, and which pass through <math>C_0, B_0</math> | ||
+ | respectively. Since <math>\triangle A_0B_0C_0, \triangle A_1B_1C_1</math> are acute | ||
+ | these lines will meet at a point <math>A</math> such that <math>B_0</math> is between <math>A, C</math> | ||
+ | and <math>C_0</math> is between <math>A, B</math>. (More about this later.) | ||
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Revision as of 16:15, 3 September 2024
Let and be any two acute-angled triangles. Consider all triangles that are similar to (so that vertices , , correspond to vertices , , , respectively) and circumscribed about triangle (where lies on , on , and on ). Of all such possible triangles, determine the one with maximum area, and construct it.
Solution
We construct a point inside s.t. , where are a permutation of . Now construct the three circles . We obtain any of the triangles circumscribed to and similar to by selecting on , then taking , and then (a quick angle chase shows that are also colinear).
We now want to maximize . Clearly, always has the same shape (i.e. all triangles are similar), so we actually want to maximize . This happens when is the diameter of . Then , so will also be the diameter of . In the same way we show that is the diameter of , so everything is maximized, as we wanted.
This solution was posted and copyrighted by grobber. The thread can be found here: [1]
Solution 2
Since all the triangles circumscribed to are similar, the one with maximum area will be the one with maximum sides, or equivalently, the one with maximum side . So we will try to maximize .
The plan is to find the value of which maximizes .
Note that for any we can construct the line through which forms the angle with . We can construct points on this line, and lines through these points which form angles with the line, and which pass through respectively. Since are acute these lines will meet at a point such that is between and is between . (More about this later.)
(Solution by pf02, September 2024)
TO BE CONTINUED. I AM SAVING MID WAY SO AS NOT TO LOSE WORK DONE SO FAR.
See Also
1967 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |