1967 IMO Problems/Problem 4
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 we will construct are are similar (by the requirement of the problem), 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 (while keeping the angles of equal to the angles of ).
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 the given angles with the line, and which pass through respectively. Since are acute, is between and these lines will meet at a point such that is between and is between .
(More about this later.)
The quantities are given. From this data, are known and constructible. We will compute in terms of and these quantities. This will be a function in the variable , and we will find the value of for which this function attains its maximum.
We will start by computing . We will use the law of sines in . We get , and a similar equality from (for ). We obtain
We can now proceed in two ways. We could use the formula for linear combination of sine functions with same period but different phase shifts (see https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Linear_combinations or https://mathworld.wolfram.com/HarmonicAdditionTheorem.html , (13)-(23)) or use calculus to find for which has its maximum value.
With the first method, we would obtain that for certain and , and we would choose such that . But we will use calculus, as a more mainstream approach. Compute the derivative and consider the equation . Use the formula for of sum of angles, and rearrange terms.
We have
Finally,
It is easy to verify that this value is valid (i.e. it can be an angle in a triangle), and it is indeed a point of maximum for (i.e. ; a geometric argument would work as well).
Now to answer the "and construct it" at the end of the statement of the problem, we will show that everything we did is constructible, rather than describe a lengthy, boring step by step construction. Recall that we already discussed that we can construct if we know . We just need to show that we can construct . We can construct differences of angles, and given an angle we can construct two segments whose ratio is the (or the , or the ) of the given angle, and vice-versa. Given three segments we can construct the segment . Thus, the expression giving is constructible.
(Solution by pf02, September 2024)
Remarks (added by pf02, September 2024)
1. In solution 2, I show where I use the condition that the triangles are assumed to be acute. The first solution does not make this clear. It seems intuitively true that the condition is not necessary. In other words, a circumscribed triangle exists, and it can be constructed even when one or both of the given triangles are right or obtuse. The condition seems to be necessary only for simplifying the proof. In the general case, we may need to rearrange the labeling of the vertices.
2. Solution 1 is elegant, even though its presentation would have benefited a lot from some editing. It gives a nice geometric insight into the problem. Solution 2 does not give much geometric insight, but it is computationally very explicit. The two solutions are so different that it is worth taking a little time to show that they are equivalent. I will outline the steps of the computation.
To show that they are equivalent, put the pictures together:
The idea from Solution 1 is to construct the circles circumscribed to and . They intersect at . Consider the angles shown on the picture. We have
The first equality follows from the law of sines in , the second follows from the equality of angles in a circle spanning the same arc, the third follows from the law of sines in . We gat . Similarly, we get from the other side of .
On the other hand, we have
Solve for and substitute in to obtain one equation in .
Now use the fact that is a diameter to write an equation between and :
Solve for in terms of and substitute in the equation in obtained before. After some straightforward computations we get exactly the equation in we had in Solution 2. The interested reader can easily work out the details, they are just straightforward algebraic computations.
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 |