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Difference between revisions of "1967 IMO Problems/Problem 1"
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We can assume <math>a > 1</math>.  Indeed, refer to Remark 2 above to
 
We can assume <math>a > 1</math>.  Indeed, refer to Remark 2 above to
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see that in this case there is nothing to prove.
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see that if <math>a \le 1</math> there is nothing to prove.
  
 
Note that instead of the statement <math>S1</math> we can consider the
 
Note that instead of the statement <math>S1</math> we can consider the
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Start like in Solution 2.
 
Start like in Solution 2.
 
We can assume <math>a > 1</math>.  Indeed, refer to Remark 2 above to
 
We can assume <math>a > 1</math>.  Indeed, refer to Remark 2 above to
−
see that in this case there is nothing to prove.
+
see that if <math>a \le 1</math> there is nothing to prove.
  
 
Note that instead of the statement <math>S1</math> we can consider the
 
Note that instead of the statement <math>S1</math> we can consider the

Revision as of 02:19, 18 September 2024

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, in Solution 2.

Also, notice that if $a \le 1$ and $\alpha \in \left( 0, \frac{\sqrt{\pi}}{2} \right)$ then $S2$ as true. 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 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 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 $\triangle BCD$.

Prob 1967 1 fig1.png

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 $D$, i.e. 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

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 $\triangle BCD$.






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

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
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