Difference between revisions of "1967 IMO Problems/Problem 4"
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equivalently, the one with maximum side <math>BC</math>. So we will try to maximize <math>BC</math>. | equivalently, the one with maximum side <math>BC</math>. So we will try to maximize <math>BC</math>. | ||
− | [[File:Prob_1967_4_fig1.png| | + | The plan is to find the value of <math>\alpha</math> which maximizes <math>BC</math>. |
+ | |||
+ | [[File:Prob_1967_4_fig1.png|600px]] | ||
+ | |||
+ | Note that for any <math>\alpha</math> we can construct the line through <math>A_0</math> which | ||
+ | forms the angle <math>\alpha</math> with <math>A_0C_0</math>. We can construct points <math>B, C</math> | ||
+ | on this line, and lines through these points which form angles | ||
+ | <math>\angle B, \angle C</math> with the line, and which pass through <math>C_0, B_0</math> | ||
+ | respectively. Since <math>\triangle A_0B_0C_0, \triangle A_1B_1C_1</math> are acute, | ||
+ | <math>A_0</math> is between <math>B, C</math> and these lines will meet at a point <math>A</math> such that | ||
+ | <math>B_0</math> is between <math>A, C</math> and <math>C_0</math> is between <math>A, B</math>. | ||
+ | |||
+ | (More about this later.) | ||
+ | |||
+ | The quantities <math>a_0, b_0, c_0, \angle B, \angle C</math> are given. From | ||
+ | this data, <math>\angle A_0, \angle B_0, \angle C_0, \angle A</math> are known and | ||
+ | constructible. We will compute <math>BC</math> in terms of <math>\angle \alpha</math> and | ||
+ | these quantities. This will be a function in the variable <math>\alpha</math>, | ||
+ | and we will find the value of <math>\alpha</math> for which this function attains | ||
+ | its maximum. | ||
+ | |||
+ | We will start by computing <math>A_0B</math>. We will use the law of sines in | ||
+ | <math>\triangle A_0C_0B</math>. We get | ||
+ | <math>\frac{A_0B}{\sin (\pi - B - \alpha)} = \frac{b_0}{\sin B}</math>, and a | ||
+ | similar equality from <math>\triangle A_0B_0C</math> (for <math>A_0C</math>). We obtain | ||
+ | |||
+ | <math>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)</math> | ||
+ | |||
+ | 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 <math>\alpha</math> for which <math>f(\alpha)</math> has its maximum value. | ||
+ | |||
+ | With the first method, we would obtain that | ||
+ | <math>f(\alpha) = D \sin (\alpha + \theta)</math> for certain <math>D</math> and <math>\theta</math>, | ||
+ | and we would choose <math>\alpha</math> such that <math>\alpha + \theta = \pi/2</math>. | ||
+ | But we will use calculus, as a more mainstream approach. Compute the | ||
+ | derivative <math>f'(\alpha)</math> and consider the equation <math>f'(\alpha) = 0</math>. | ||
+ | Use the formula for <math>\cos</math> of sum of angles, and rearrange terms. | ||
+ | |||
+ | We have | ||
+ | |||
+ | <math>\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 ]</math> | ||
+ | |||
+ | Finally, | ||
+ | |||
+ | <math>\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)}</math> | ||
+ | |||
+ | It is easy to verify that this value is valid, and it is indeed | ||
+ | a point of maximum for <math>f(\alpha)</math>. | ||
+ | |||
+ | 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 <math>\triangle ABC</math> if | ||
+ | we know <math>\alpha</math>. We just need to show that we can construct | ||
+ | <math>\alpha</math>. We can construct differences of angles, and given an | ||
+ | angle we can construct two segments whose ratio is the <math>\sin</math> | ||
+ | (or the <math>\cos</math>, or the <math>\tan</math>) of the given angle, and vice-versa. | ||
+ | Given three segments <math>x, y, z</math> we can construct the segment | ||
+ | <math>x \cdot \frac{y}{z}</math>. Thus, the expression giving <math>\alpha</math> | ||
+ | is constructible. | ||
+ | |||
+ | (Solution by pf02, September 2024) | ||
+ | |||
+ | |||
+ | ==(Remarks (added by pf02, September 2024)== | ||
+ | |||
+ | |||
+ | |||
+ | |||
− | |||
TO BE CONTINUED. I AM SAVING MID WAY SO AS NOT TO LOSE WORK DONE SO FAR. | TO BE CONTINUED. I AM SAVING MID WAY SO AS NOT TO LOSE WORK DONE SO FAR. |
Latest revision as of 21:00, 3 September 2024
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
are similar, 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
.
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 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, and it is indeed
a point of maximum for .
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 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)
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 |