Difference between revisions of "2019 AIME II Problems/Problem 1"
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Extend <math>AB</math> to form a right triangle with legs <math>6</math> and <math>8</math> such that <math>AD</math> is the hypotenuse and connect the points <math>CD</math> so | Extend <math>AB</math> to form a right triangle with legs <math>6</math> and <math>8</math> such that <math>AD</math> is the hypotenuse and connect the points <math>CD</math> so | ||
− | that you have a rectangle. (We know that <math>\triangle ADE</math> is a <math>6-8-10</math>, since <math>\triangle DEB</math> is an <math>8-15-17</math>.) The base <math>CD</math> of the rectangle will be <math>9+6+6=21</math>. Now, let <math> | + | that you have a rectangle. (We know that <math>\triangle ADE</math> is a <math>6-8-10</math>, since <math>\triangle DEB</math> is an <math>8-15-17</math>.) The base <math>CD</math> of the rectangle will be <math>9+6+6=21</math>. Now, let <math>O</math> be the intersection of <math>BD</math> and <math>AC</math>. This means that <math>\triangle ABO</math> and <math>\triangle DCO</math> are with ratio <math>\frac{21}{9}=\frac73</math>. Set up a proportion, knowing that the two heights add up to 8. We will let <math>y</math> be the height from <math>O</math> to <math>DC</math>, and <math>x</math> be the height of <math>\triangle ABO</math>. |
<cmath>\frac{7}{3}=\frac{y}{x}</cmath> | <cmath>\frac{7}{3}=\frac{y}{x}</cmath> | ||
<cmath>\frac{7}{3}=\frac{8-x}{x}</cmath> | <cmath>\frac{7}{3}=\frac{8-x}{x}</cmath> |
Revision as of 17:08, 26 December 2020
Contents
[hide]Problem
Two different points, and
, lie on the same side of line
so that
and
are congruent with
,
, and
. The intersection of these two triangular regions has area
, where
and
are relatively prime positive integers. Find
.
Solution
- Diagram by Brendanb4321
Extend to form a right triangle with legs
and
such that
is the hypotenuse and connect the points
so
that you have a rectangle. (We know that
is a
, since
is an
.) The base
of the rectangle will be
. Now, let
be the intersection of
and
. This means that
and
are with ratio
. Set up a proportion, knowing that the two heights add up to 8. We will let
be the height from
to
, and
be the height of
.
This means that the area is . This gets us
-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers
Solution 2
Using the diagram in Solution 1, let be the intersection of
and
. We can see that angle
is in both
and
. Since
and
are congruent by AAS, we can then state
and
. It follows that
and
. We can now state that the area of
is the area of
the area of
. Using Heron's formula, we compute the area of
. Using the Law of Cosines on angle
, we obtain
(For convenience, we're not going to simplify.)
Applying the Law of Cosines on yields
This means
. Next, apply Heron's formula to get the area of
, which equals
after simplifying. Subtracting the area of
from the area of
yields the area of
, which is
, giving us our answer, which is
-Solution by flobszemathguy
Solution 3 (Very quick)
- Diagram by Brendanb4321 extended by Duoquinquagintillion
Begin with the first step of solution 1, seeing is the hypotenuse of a
triangle and calling the intersection of
and
point
. Next, notice
is the hypotenuse of an
triangle. Drop an altitude from
with length
, so the other leg of the new triangle formed has length
. Notice we have formed similar triangles, and we can solve for
.
So has area
And
- Solution by Duoquinquagintillion
Solution 4
Let . By Law of Cosines,
And
- by Mathdummy
Solution 5
Because and
, quadrilateral
is cyclic. So, Ptolemy's theorem tells us that
From here, there are many ways to finish which have been listed above. If we let , then
Using Heron's formula on , we see that
Thus, our answer is . ~a.y.711
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.