Difference between revisions of "2019 AIME I Problems/Problem 11"
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~[[User:Icematrix2|icematrix2]] | ~[[User:Icematrix2|icematrix2]] | ||
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+ | ==Solution 5 (Standard geometry)== | ||
+ | [[File:2019 AIME I 11a.png|500px|right]] | ||
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+ | Some notation first: let <math>M</math> be the midpoint <math>BC, BM = a, AB= BC = b, s = b+a</math> be the semiperimeter, <math>r</math> be the inradius.<math>I_A, I_B</math> be excenters, <math>r_A, r_B</math> be exradius, <math>R</math> be radius <math>\omega.</math> | ||
==Video Solution (On the Spot STEM)== | ==Video Solution (On the Spot STEM)== |
Revision as of 16:15, 11 June 2022
Contents
Problem
In , the sides have integer lengths and
. Circle
has its center at the incenter of
. An excircle of
is a circle in the exterior of
that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to
is internally tangent to
, and the other two excircles are both externally tangent to
. Find the minimum possible value of the perimeter of
.
Solution 1
Let the tangent circle be . Some notation first: let
,
,
be the semiperimeter,
, and
be the inradius. Intuition tells us that the radius of
is
(using the exradius formula). However, the sum of the radius of
and
is equivalent to the distance between the incenter and the the
excenter. Denote the B excenter as
and the incenter as
.
Lemma:
We draw the circumcircle of
. Let the angle bisector of
hit the circumcircle at a second point
. By the incenter-excenter lemma,
. Let this distance be
. Ptolemy's theorem on
gives us
Again, by the incenter-excenter lemma,
so
as desired.
Using this gives us the following equation:
Motivated by the
and
, we make the following substitution:
This changes things quite a bit. Here's what we can get from it:
It is known (easily proved with Heron's and
) that
Using this, we can also find
: let the midpoint of
be
. Using Pythagorean's Theorem on
,
We now look at the RHS of the main equation:
Cancelling some terms, we have
Squaring,
Expanding and moving terms around gives
Reverse substituting,
Clearly the smallest solution is
and
, so our answer is
-franchester
Solution 2 (Lots of Pythagorean Theorem)
First, assume and
. The triangle can be scaled later if necessary. Let
be the incenter and let
be the inradius. Let the points at which the incircle intersects
,
, and
be denoted
,
, and
, respectively.
Next, we calculate in terms of
. Note the right triangle formed by
,
, and
. The length
is equal to
. Using the Pythagorean Theorem, the length
is
, so the length
is
. Note that
is half of
, and by symmetry caused by the incircle,
and
, so
. Applying the Pythagorean Theorem to
, we get
Expanding yields
which can be simplified to
Dividing by
and then squaring results in
and isolating
gets us
so
.
We then calculate the radius of the excircle tangent to . We denote the center of the excircle
and the radius
.
Consider the quadrilateral formed by ,
,
, and the point at which the excircle intersects the extension of
, which we denote
. By symmetry caused by the excircle,
, so
.
Note that triangles and
are congruent, and
and
are also congruent. Denoting the measure of angles
and
measure
and the measure of angles
and
measure
, straight angle
, so
. This means that angle
is a right angle, so it forms a right triangle.
Setting the base of the right triangle to , the height is
and the base consists of
and
. Triangles
and
are similar to
, so
, or
. This makes
the reciprocal of
, so
.
Circle 's radius can be expressed by the distance from the incenter
to the bottom of the excircle with center
. This length is equal to
, or
. Denote this value
.
Finally, we calculate the distance from the incenter to the closest point on the excircle tangent to
, which forms another radius of circle
and is equal to
. We denote the center of the excircle
and the radius
. We also denote the points where the excircle intersects
and the extension of
using
and
, respectively. In order to calculate the distance, we must find the distance between
and
and subtract off the radius
.
We first must calculate the radius of the excircle. Because the excircle is tangent to both and the extension of
, its center must lie on the angle bisector formed by the two lines, which is parallel to
. This means that the distance from
to
is equal to the length of
, so the radius is also
.
Next, we find the length of . We can do this by forming the right triangle
. The length of leg
is equal to
minus
, or
. In order to calculate the length of leg
, note that right triangles
and
are congruent, as
and
share a length of
, and angles
and
add up to the right angle
. This means that
.
Using Pythagorean Theorem, we get
Bringing back
and substituting in some values, the equation becomes
Rearranging and squaring both sides gets
Distributing both sides yields
Canceling terms results in
Since
We can further simplify to
Substituting out
gets
which when distributed yields
After some canceling, distributing, and rearranging, we obtain
Multiplying both sides by
results in
which can be rearranged into
and factored into
This means that
equals
or
, and since a side length of
cannot exist,
.
As a result, the triangle must have sides in the ratio of . Since the triangle must have integer side lengths, and these values share no common factors greater than
, the triangle with the smallest possible perimeter under these restrictions has a perimeter of
. ~emerald_block
Solution 3 (Various Techniques)
Before we start thinking about the problem, let’s draw it out;
For the sake of space, I've drawn only 2 of the 3 excircles because the third one looks the same as the second large one because the triangle is isosceles. By the incenter-excenter lemma, and
are collinear,
is the tangent of circle
to
,
is the tangent of that circle to the extension of
, and
is the tangent of the circle to the extension of
. The interesting part of the diagram is circle
, which is internally tangent to circle
yet externally tangent to circle
. Therefore, perhaps we can relate the radius of this circle to the semiperimeter of triangle
.
We can see that the radius of circle is
using the incenter and A-excenter of our main triangle. This radius is also equal to
from the incenter and B-excenter of our triangle. Thus, we can solve for each of these separately in terms of the lengths of the triangle and set them equal to each other to form an equation.
To find the left hand side of the equation, we have to first find and
. Let
and
. Then since the perimeter of the triangle is
, the semiperimeter is
.
Now let's take a look at triangle . Because
is the angle bisector of
, by the angle bisector theorem,
. Rearranging, we get
.
Take a look at triangle .
,
(angle bisector theorem), and
. Now let's analyze triangle
.
,
, and
. Since
and
, triangle
and
are similar by AA. Then
. Now, solving yields
.
Finally, the left hand side of our equation is
Now let's look at triangle . How will we find
? Let's first try to find
and
in terms of the lengths of the triangle. We recognize:
. We really want to have
instead of
, and
looks very similar in length to
, so let's try to prove that they are equal.
, so we can try to add these two and see if we get anything interesting. We have:
, which is our perimeter. Thus,
.
Triangle is similar to triangle
by AA, and we know that
, and
, so thus
. Thus, the height of this triangle is
by similarity ratios, the same height as vertex
. By the Pythagorean Theorem,
and by similarity ratios,
. Finally,
, and thus the right hand side of our equation is
.
Setting the two equal, we have
Multiplying both sides by we have
From here, let arbitrarily; note that we can always scale this value to fit the requirements later. Thus our equation is
. Now since
, we can plug into our equation:
. Remembering
;
because the side lengths have to be positive numbers. Furthermore, because our values for
and
are relatively prime, we don't have to scale down our triangle further, and we are done. Therefore, our answer is
~KingRavi
Solution 4 (Not that hard construction)
Notice that the -excircle would have to be very small to fit the property that it is internally tangent to
and the other two excircles are both externally tangent, given that circle
's centre is at the incenter of
. If
, we see that
must be somewhere in the
to
range. If we test
by construction, we notice the
-excircle is too big for it to be internally tangent to
while the other two are externally tangent. This means we should test
or
next. I actually did this and found that
worked, so the answer is
. Note that
cannot be
because then
would have to be
which is not an integer.
Solution 5 (Standard geometry)
Some notation first: let be the midpoint
be the semiperimeter,
be the inradius.
be excenters,
be exradius,
be radius
Video Solution (On the Spot STEM)
This solution is the video solution for Solution 3 - not posted by ~KingRavi
https://www.youtube.com/watch?v=zKHwTJBhKdM
Video Solution 2 (More concise)
https://www.youtube.com/watch?v=ldr4yi3t6hQ
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.