1967 IMO Problems/Problem 1
Let be a parallelogram with side lengths , , and with . If is acute, prove that the four circles of radius with centers , , , cover the parallelogram if and only if
Solution
To start our proof we draw a parallelogram with the requested sides. We notice that by drawing the circles with centers A, B, C, D that the length of must not exceed 2 (the radius for each circle) or the circles will not meet and thus not cover the parallelogram.
To prove our conjecture we draw a parallelogram with and draw a segment so that
This is the parallelogram which we claim has the maximum length on and the highest value on any one angle.
We now have two triangles inside a parallelogram with lengths and , being segment . Using the Pythagorean theorem we conclude:
Using trigonometric functions we can compute:
Notice that by applying the and functions, we can conclude that our angle
To conclude our proof we make sure that our values match the required values for maximum length of
Notice that as decreases, the value of (1) increases beyond 2. We can prove this using the law of sines. Similarly as increases, the value of (1) decreases below 2, confirming that (1) is only implied when is acute.
--Bjarnidk 02:16, 17 May 2013 (EDT)
Remarks (added by pf02, September 2024)
. I am sorry to be so harshly critical, but the solution above is deeply flawed. Not only it has errors, but the logic is flawed.
It shows that when the parallelogram is covered by the circles of radius centered at , and the inequality in the problem is true. (Even this is incomplete, while giving too many, unnecessary details.) (Note that this is not a case which satisfies the conditions of the problem since is right, not acute.)
In the last two lines it gives some reasoning about other values of which is incomprehensible to this reader.
In one short sentence: this is not a solution.
. The problem itself is mildly flawed. To see this, denote the following two statements:
S1: The parallelogram is covered by the four circles of radius centered at .
S2: We have .
The problem says that if is acute, and are equivalent, i.e. they imply each other.
Notice that can be rewritten as .
Now notice that if then S1 is obviously true.
Also, notice that if and then as true. Indeed , so is on this interval, so the right hand side of is .
We see that if and is acute, both and are true. We can not say that one implies the other in the usual meaning of the word "imply": the two statements just happen to be both true.
If we take then the problem is a genuine problem, and there is something to prove.
. As we will see in the proofs I give below, we don't need to know that is acute. All we need is that is acute.
In fact, it is possible to modify to a statement similar to so that and are equivalent without any assumption on . I will not go into this, I will just give a hint: Denote . If is acute, is obtuse, and we can easily reformulate in terms of .
. Below, I will give two solutions. Solution 2 is one I carried out myself. Solution 3 is inspired by an idea by feliz shown on the web page https://artofproblemsolving.com/community/c6h21154p137323 The author calls it a solution, but it is so confused that it can hardly be called a solution. The idea though is good and nice, and it yields a nice solution.
Solution 2
TO BE CONTINUED. I AM SAVING MID WAY SO AS NOT TO LOSE WORK DONE SO FAR.
A solution can also be found here [1]
See Also
1967 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |