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Difference between revisions of "1967 IMO Problems/Problem 4"

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Line 91: Line 91:
 
==(Remarks (added 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 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.
  
 +
To show that they are equivalent, put the pictures together:
 +
 +
[[File:Prob_1967_4_fig2.png|600px]]
  
  

Revision as of 23:30, 3 September 2024

Let $A_0B_0C_0$ and $A_1B_1C_1$ be any two acute-angled triangles. Consider all triangles $ABC$ that are similar to $\triangle A_1B_1C_1$ (so that vertices $A_1$, $B_1$, $C_1$ correspond to vertices $A$, $B$, $C$, respectively) and circumscribed about triangle $A_0B_0C_0$ (where $A_0$ lies on $BC$, $B_0$ on $CA$, and $AC_0$ on $AB$). Of all such possible triangles, determine the one with maximum area, and construct it.


Solution

We construct a point $P$ inside $A_0B_0C_0$ s.t. $\angle X_0PY_0=\pi-\angle X_1Z_1Y_1$, where $X,Y,Z$ are a permutation of $A,B,C$. Now construct the three circles $\mathcal C_A=(B_0PC_0),\mathcal C_B=(C_0PA_0),\mathcal C_C=(A_0PB_0)$. We obtain any of the triangles $ABC$ circumscribed to $A_0B_0C_0$ and similar to $A_1B_1C_1$ by selecting $A$ on $\mathcal C_A$, then taking $B= AB_0\cap \mathcal C_C$, and then $B=CA_0\cap\mathcal C_B$ (a quick angle chase shows that $B,C_0,A$ are also colinear).

We now want to maximize $BC$. Clearly, $PBC$ always has the same shape (i.e. all triangles $PBC$ are similar), so we actually want to maximize $PB$. This happens when $PB$ is the diameter of $\mathcal C_B$. Then $PA_0\perp BC$, so $PC$ will also be the diameter of $\mathcal C_C$. In the same way we show that $PA$ is the diameter of $\mathcal C_A$, 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 $\triangle ABC$ circumscribed to $\triangle A_0B_0C_0$ are similar, the one with maximum area will be the one with maximum sides, or equivalently, the one with maximum side $BC$. So we will try to maximize $BC$.

The plan is to find the value of $\alpha$ which maximizes $BC$.

Prob 1967 4 fig1.png

Note that for any $\alpha$ we can construct the line through $A_0$ which forms the angle $\alpha$ with $A_0C_0$. We can construct points $B, C$ on this line, and lines through these points which form angles $\angle B, \angle C$ with the line, and which pass through $C_0, B_0$ respectively. Since $\triangle A_0B_0C_0, \triangle A_1B_1C_1$ are acute, $A_0$ is between $B, C$ and these lines will meet at a point $A$ such that $B_0$ is between $A, C$ and $C_0$ is between $A, B$.

(More about this later.)

The quantities $a_0, b_0, c_0, \angle B, \angle C$ are given. From this data, $\angle A_0, \angle B_0, \angle C_0, \angle A$ are known and constructible. We will compute $BC$ in terms of $\angle \alpha$ and these quantities. This will be a function in the variable $\alpha$, and we will find the value of $\alpha$ for which this function attains its maximum.

We will start by computing $A_0B$. We will use the law of sines in $\triangle A_0C_0B$. We get $\frac{A_0B}{\sin (\pi - B - \alpha)} = \frac{b_0}{\sin B}$, and a similar equality from $\triangle A_0B_0C$ (for $A_0C$). We obtain

$BC = A_0B + A_0C = \frac{b_0}{\sin B} \sin (B + \alpha) + \frac{c_0}{\sin C} \sin (A_0 - C + \alpha) = f(\alpha)$

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 $\alpha$ for which $f(\alpha)$ has its maximum value.

With the first method, we would obtain that $f(\alpha) = D \sin (\alpha + \theta)$ for certain $D$ and $\theta$, and we would choose $\alpha$ such that $\alpha + \theta = \pi/2$. But we will use calculus, as a more mainstream approach. Compute the derivative $f'(\alpha)$ and consider the equation $f'(\alpha) = 0$. Use the formula for $\cos$ of sum of angles, and rearrange terms.

We have

$\cos \alpha \cdot \left [ \frac{b_0}{\sin B} \cdot \cos B + \frac{c_0}{\sin C} \cdot \cos (A_0 - C) \right ]= \sin \alpha \cdot \left [ \frac{b_0}{\sin B} \cdot \sin B + \frac{c_0}{\sin C} \cdot \sin (A_0 - C) \right ]$

Finally,

$\alpha = \arctan \frac{b_0 \cos B \sin C + c_0 \sin B \cos (A_0 - C)} {b_0 \sin B \sin C + c_0 \sin B \sin (A_0 - C)}$

It is easy to verify that this value is valid, and it is indeed a point of maximum for $f(\alpha)$.

Now to answer the "and construct it" part of the problem, we will show that everything we did is constructible, rather than give a lengthy, boring step by step construction. Recall that we already discussed that we can construct $\triangle ABC$ if we know $\alpha$. We just need to show that we can construct $\alpha$. We can construct differences of angles, and given an angle we can construct two segments whose ratio is the $\sin$ (or the $\cos$, or the $\tan$) of the given angle, and vice-versa. Given three segments $x, y, z$ we can construct the segment $x \cdot \frac{y}{z}$. Thus, the expression giving $\alpha$ 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 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.

To show that they are equivalent, put the pictures together:

Prob 1967 4 fig2.png



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
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