1967 IMO Problems/Problem 1

Let $ABCD$ be a parallelogram with side lengths $AB = a$ , $AD = 1$ , and with $\angle BAD = \alpha$ . If $\Delta ABD$ is acute, prove that the four circles of radius $1$ with centers $A$ , $B$ , $C$ , $D$ cover the parallelogram if and only if

$a\leq \cos \alpha+\sqrt{3}\sin \alpha$ $\ \ \ \ \ \ \ \ \ \ (1)$

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 $a$ 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 $a=2$ and draw a segment $DB$ so that $\angle ADB=90^{\circ}$

This is the parallelogram which we claim has the maximum length on $a$ and the highest value on any one angle.

We now have two triangles inside a parallelogram with lengths $1, 2$ and $x$ , $x$ being segment $DB$ . Using the Pythagorean theorem we conclude:

$1^2+x^2=2^2\\x=\sqrt{3}$

Using trigonometric functions we can compute:

$cos\alpha=\frac{1}{2}\\sin\alpha=\frac{\sqrt{3}}{2}$

Notice that by applying the $arcsine$ and $arccos$ functions, we can conclude that our angle $\alpha=60^{\circ}$

To conclude our proof we make sure that our values match the required values for maximum length of $a$

$a\leq\cos\alpha+\sqrt{3}\sin\alpha\\\\a\leq\frac{1}{2}+\sqrt{3}\cdot \frac{\sqrt{3}}{2}\\\\a\leq 2$

Notice that as $\angle\alpha$ decreases, the value of (1) increases beyond 2. We can prove this using the law of sines. Similarly as $\angle\alpha$ increases, the value of (1) decreases below 2, confirming that (1) is only implied when $\Delta ABD$ is acute.

--Bjarnidk 02:16, 17 May 2013 (EDT)

Remarks (added by pf02, September 2024)

$\mathbf{Remark\ 1}$ . 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 $a = 2, \alpha = \frac{\pi}{3}$ the parallelogram is covered by the circles of radius $1$ centered at $A, B, C, D$ , 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 $\triangle ABD$ is right, not acute.)

In the last two lines it gives some reasoning about other values of $\alpha$ which is incomprehensible to this reader.

In one short sentence: this is not a solution.

$\mathbf{Remark\ 2}$ . The problem itself is mildly flawed. To see this, denote $S1, S2$ the following two statements:

S1: The parallelogram $ABCD$ is covered by the four circles of radius $1$ centered at $A, B, C, D$ .

S2: We have $a \le \cos \alpha + \sqrt{3} \sin \alpha$ .

The problem says that if $\triangle ABD$ is acute, $S1$ and $S2$ are equivalent, i.e. they imply each other.

Notice that $S2$ can be rewritten as $a \le 2 \cos \left( \alpha - \frac{\sqrt{\pi}}{3} \right)$ .

Now notice that if $a \le 1$ then S1 is obviously true. See the picture below:

Also, notice that if $a \le 1$ and $\alpha \in \left( 0, \frac{\sqrt{\pi}}{2} \right)$ then $S2$ is true as well. Indeed $\left( \alpha - \frac{\sqrt{\pi}}{3} \right) \in \left( -\frac{\sqrt{\pi}}{3}, \frac{\sqrt{\pi}}{6} \right)$ , so $\cos$ is $> \frac{1}{2}$ on this interval, so the right hand side of $S2$ is $> 1 \ge a$ .

We see that if $a \le 1$ and $\triangle ABD$ is acute, both $S1$ and $S2$ 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 $a > 1$ then the problem is a genuine problem, and there is something to prove.

$\mathbf{Remark\ 3}$ . In the proofs I give below, we will see where we need that $\triangle ABD$ is acute. We will see that $\alpha < \frac{\pi}{2}$ is needed for the technicalities of the proof. The fact that $\angle ADB$ is acute will be needed at one crucial point in the proof.

In fact, it is possible to modify $S2$ to a statement $S3$ similar to $S2$ so that $S1$ and $S3$ are equivalent without any assumption on $\alpha$ . I will not go into this, I will just give a hint: Denote $\beta = \angle ABC$ . If $\alpha$ is acute, $\beta$ is obtuse, and we can easily reformulate $S2$ in terms of $\beta$ .

$\mathbf{Remark\ 4}$ . Below, I will give two solutions. Solution 2 is one I carried out myself and relies on a straightforward computation. Solution 3 (it happens to be similar to Solution 2) is inspired by an idea by feliz shown on the web page https://artofproblemsolving.com/community/c6h21154p137323 The author calls their text a solution, but it is quite confused, so I would not call it a good solution. The idea though is good and nice, and it yields a nice solution.

Solution 2

We can assume $a > 1$ . Indeed, refer to Remark 2 above to see that if $a \le 1$ there is nothing to prove.

Note that instead of the statement $S1$ we can consider the following statement $S1'$ :

$S1'$ : the circles of radius $1$ centered at $A, B, D$ cover $\triangle ABD$ .

This is equivalent to $S1$ because of the symmetry between $\triangle ABD$ and $\triangle BCD$ .

Let $F$ be the intersection above $AB$ of the circles of radius $1$ centered at $A, B$ . The three circles cover $\triangle ABD$ if an only if $F$ is inside the circle of radius 1 centered at $D$ .

This needs an explanation: Let $H$ be the midpoint of $BD$ , and consider $\triangle FHD$ . All the vertices of this triangle are in the circle centered at $D$ , so the whole triangle is inside this circle. It is obvious that $\triangle FHB$ is inside the circle centered at $B$ , and that $\triangle FAD, \triangle FAB$ are inside the circles centered at $A, B$ .

We will now use that $F$ is inside the circle of radius 1 centered at $D$ if an only if $DF \le 1$ .

The plan is to calculate $DF$ in terms of $a, \alpha$ and impose this condition. Let $FG \perp AB$ , $DE \perp AB$ and $FF' \parallel GE$ . From the right triangle $\triangle AFG$ we have $FG = \sqrt{1 - \left( \frac{a}{2} \right)^2} = \frac{\sqrt{4 - a^2}}{2}$ . From the right triangle $\triangle DFF'$ we have

$DF = \sqrt{(DF')^2 + (FF')^2} = \sqrt{(DE - FG)^2 + (AG - AE)^2} = \sqrt{\left( \sin \alpha - \frac{\sqrt{4 - a^2}}{2} \right)^2 + \left( \frac{a}{2} - \cos \alpha \right)^2}$

(Note that here we used the fact that $\alpha$ is acute. These equalities would look slightly differently otherwise.)

Now look at the condition $DF \le 1$ , or equivalently $DF^2 \le 1$ . Making all the computations and simplifications, we have $\sqrt{4 - a^2} \sin \alpha \ge 1 - a \cos \alpha$ .

Now I would like to square both sides. In order to get an equivalent inequality, we need to know that $1 - a \cos \alpha \ge 0$ . This follows from the fact that $\angle ADB$ is acute. Indeed, denote $\angle ADB = \beta$ . From the law of sines in $\triangle ADB$ we have $\frac{1}{\sin \angle ABD} = \frac{a}{\sin \beta}$ . Successively this becomes $\frac{1}{\sin (\pi - \alpha - \beta)} = \frac{a}{\sin \beta}$ or $\frac{1}{\sin (\alpha + \beta)} = \frac{a}{\sin \beta}$ or $(1 - a \cos \alpha) \sin \beta = a \sin \alpha \cos \beta$ . From here we see that $\beta < \frac{\pi}{2}$ implies the right hand side is positive, so $1 - a \cos \alpha > 0$ .

Going back to our inequality, we can square both sides, and after rearranging terms we get that $DF \le 1$ if and only if

$a^2 - 2a \cos \alpha + (1 - 4 \sin^2 \alpha) \le 0$ .

View this as an equation of degree $2$ in $a$ . The value of the polynomial in $a$ is $\le 0$ when $a$ is between its solutions, that is

$\cos \alpha - \sqrt{3} \sin \alpha \le a \le \cos \alpha + \sqrt{3} \sin \alpha$ .

Note that $\cos \alpha - \sqrt{3} \sin \alpha = 2 \cos \left( \alpha + \frac{\pi}{3} \right)$ . If $\alpha \in \left( 0, \frac{\pi}{2} \right)$ then $\left( \alpha + \frac{\pi}{3} \right) \in \left( \frac{\pi}{3}, \frac{5\pi}{6} \right)$ , and it follows that $2 \cos \left( \alpha + \frac{\pi}{3} \right) \le 1$ .

On the other hand, remember that we are in the case $a > 1$ , so the left inequality is always true. It follows that

$DF \le 1$ (i.e. the three circles of radius $1$ centered at $A, B, D$ cover $\triangle ABD$ ) if an only if $a \le \cos \alpha + \sqrt{3} \sin \alpha$ .

(Solution by pf02, September 2024)

Solution 3

This solution is very similar to Solution 2, except that we choose another point instead of $F$ . This will in fact simplify the proof. Start like in Solution 2.

We can assume $a > 1$ . Indeed, refer to Remark 2 above to see that if $a \le 1$ there is nothing to prove.

Note that instead of the statement $S1$ we can consider the following statement $S1'$ :

$S1'$ : the circles of radius $1$ centered at $A, B, D$ cover $\triangle ABD$ .

This is equivalent to $S1$ because of the symmetry between $\triangle ABD$ and $\triangle BCD$ .

Let $O$ be the center of the circle circumscribed to $\triangle ABD$ . Let $M, N, P$ , be the midpoints of $AB, AD, BD$ . The three circles cover $\triangle ABD$ if an only if $O$ is inside the circle of radius 1 centered at $D$ .

This needs an explanation. In fact, since $OA = OB = OD$ , the point $O$ is inside or on the circle centered at $D$ if and only if $OD \le 1$ , if and only if $O$ is in or inside the circles centered at $A, B, D$ . Since $OP \perp BD$ , $DP < OD, DB < OB$ , so the triangles $\triangle OPD, \triangle OPB$ are inside the circles centered at $D, B$ respectively. By drawing $OA, OP$ we can easily verify that the whole triangle is inside the circles centered at $A, D, B$ .

Note that in the above argument we used that $O$ is inside $\triangle ABD$ , which is true because the triangle is acute.

Denote $R$ the radius of the circle circumscribed to $\triangle ABD$ . From the law of sines, we have $\frac{BD}{\sin \alpha} = 2R$ , and from the law of cosines we have $BD^2 = 1 + a^2 -2a \cos \alpha$ . So $R = OD \le 1$ translates to $\frac{\sqrt{1 + a^2 -2a \cos \alpha}}{2 \sin \alpha} \le 1$

Since $\sin \alpha > 0$ , we can simplify this inequality, and get

$a^2 - 2a \cos \alpha + (1 - 4 \sin^2 \alpha) \le 0$ .

But this is exactly the inequality we encountered in Solution 2, so proving that this is equivalent to

$a \le \cos \alpha + \sqrt{3} \sin \alpha$

is identical to what has been done above.

(Solution based on an idea by feliz; see link in Remark 4, or below.)

A solution can also be found here [1]

1967 IMO (Problems) • Resources
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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1967 IMO Problems/Problem 1

Contents

Solution

Remarks (added by pf02, September 2024)

Solution 2

Solution 3

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