Difference between revisions of "2022 AMC 10A Problems/Problem 15"
MRENTHUSIASM (talk | contribs) m (→Solution 3 (if you didn't realize the diagonal was the diameter)) |
|||
Line 61: | Line 61: | ||
== Solution 3 (if you didn't realize the diagonal was the diameter) == | == Solution 3 (if you didn't realize the diagonal was the diameter) == | ||
− | We can observe that this quadrilateral is actually made of two right triangles: <math>\triangle CDA</math> has a <math>3-4-5</math> ratio in the side lengths, and <math>\triangle ABC</math> is a <math>7-24-25</math> triangle. | + | We can observe that this quadrilateral is actually made of two right triangles: <math>\triangle CDA</math> has a <math>3 \text{-} 4 \text{-} 5</math> ratio in the side lengths, and <math>\triangle ABC</math> is a <math>7 \text{-} 24 \text{-} 25</math> triangle. |
Next, we can choose one of these triangles and use the circumradius formula to find the radius. Let's choose the <math>15-20-25</math> triangle. The area of the triangle is equal to the product of the side lengths divided by <math>4</math> times the circumradius. Therefore, <math>150 = \frac{15\cdot20\cdot25}{4r}</math>. Solving this simple algebraic equation gives us <math>r = \frac{25}{2}</math>. | Next, we can choose one of these triangles and use the circumradius formula to find the radius. Let's choose the <math>15-20-25</math> triangle. The area of the triangle is equal to the product of the side lengths divided by <math>4</math> times the circumradius. Therefore, <math>150 = \frac{15\cdot20\cdot25}{4r}</math>. Solving this simple algebraic equation gives us <math>r = \frac{25}{2}</math>. | ||
Line 70: | Line 70: | ||
~[[OrenSH|orenbad]] | ~[[OrenSH|orenbad]] | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Like in Solution 3, <math>\overline{AC}</math> splits the quadrilateral into two right triangles: <math>\triangle ABC</math> and <math>\triangle ADC</math>, which are <math>7-24-25</math> and <math>15-20-25</math> triangles, respectively. So, the area of the quadrilateral is the sum of the areas of the two triangles. <math>\frac{7 \cdot 24}{2} + \frac{15 \cdot 20}{2} = 234</math>. The diameter is <math>25</math>, so the radius is <math>\frac{25}{2}</math>. Hence, the shaded area is <math>\frac{25}{2}^2 \pi - 234 = \frac{625}{4} \pi - 234 = \frac{625 \pi - 936}{4} \Rightarrow \boxed{\textbf{(D) }1565}</math>. | ||
+ | |||
+ | ~MrThinker | ||
== Video Solution == | == Video Solution == |
Revision as of 23:19, 29 November 2022
Contents
Problem
Quadrilateral with side lengths
is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form
where
and
are positive integers such that
and
have no common prime factor. What is
Diagram
~MRENTHUSIASM
Solution 1 (Inscribed Angle Theorem)
Opposite angles of every cyclic quadrilateral are supplementary, so
We claim that
We can prove it by contradiction:
- If
then
and
are both acute angles. This arrives at a contradiction.
- If
then
and
are both obtuse angles. This arrives at a contradiction.
By the Inscribed Angle Theorem, we conclude that is the diameter of the circle. So, the radius of the circle is
The area of the requested region is
Therefore, the answer is
~MRENTHUSIASM
Solution 2 (Brahmagupta‘s Formula)
When we look at the side lengths of the quadrilateral we see and
which screams out
because of Pythagorean triplets. As a result, we can draw a line through points
and
to make a diameter of
. Since the diameter is
we can see the area of the circle is just
from the formula of the area of the circle with just a diameter.
- Then we can use Brahmagupta Formula,
where
are side lengths, and s is semi-perimeter to find the area of the quadrilateral.
If we just plug the values in, we get which is
. So now the area of the region we are trying to find is
or it can be written as
Therefore, the answer is
~Gdking
Solution 3 (if you didn't realize the diagonal was the diameter)
We can observe that this quadrilateral is actually made of two right triangles: has a
ratio in the side lengths, and
is a
triangle.
Next, we can choose one of these triangles and use the circumradius formula to find the radius. Let's choose the triangle. The area of the triangle is equal to the product of the side lengths divided by
times the circumradius. Therefore,
. Solving this simple algebraic equation gives us
.
Plugging in the values, we have . Rewriting this gives us
.
Therefore, adding these values gets us
Solution 4
Like in Solution 3, splits the quadrilateral into two right triangles:
and
, which are
and
triangles, respectively. So, the area of the quadrilateral is the sum of the areas of the two triangles.
. The diameter is
, so the radius is
. Hence, the shaded area is
.
~MrThinker
Video Solution
- Whiz
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.