1967 IMO Problems/Problem 4
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
we will construct are are similar (by the requirement of the problem), 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
(while keeping the
angles of
equal to the angles of
).
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 the given 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 (i.e. it can be an angle
in a triangle), and it is indeed a point of maximum for
(i.e.
; a geometric argument would work as well).
Now to answer the "and construct it" at the end of the statement of the
problem, we will show that everything we did is constructible, rather
than describe 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)
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 given 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. I will outline the steps of the computation.
To show that they are equivalent, put the pictures together:
The idea from Solution 1 is to construct the circles circumscribed to
and
. They intersect at
.
Consider the angles
shown on the picture. We have
The first equality follows from the law of sines in ,
the second follows from the equality of angles in a circle spanning the
same arc, the third follows from the law of sines in
.
We gat
. Similarly, we get
from the other side of
.
On the other hand, we have
Solve for and substitute in
to obtain one equation in
.
Now use the fact that is a diameter to write an equation between
and
:
Solve for in terms of
and substitute in the equation
in
obtained before. After some straightforward computations
we get exactly the equation in
we had in Solution 2. The
interested reader can easily work out the details, they are just
straightforward algebraic computations.
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 |