2020 AMC 10A Problems/Problem 20
- The following problem is from both the 2020 AMC 12A #18 and 2020 AMC 10A #20, so both problems redirect to this page.
Contents
Problem
Quadrilateral satisfies
and
Diagonals
and
intersect at point
and
What is the area of quadrilateral
Solution 1 (Just Drop An Altitude)
It's crucial to draw a good diagram for this one. Since and
, we get
. Now we need to find
to get the area of the whole quadrilateral. Drop an altitude from
to
and call the point of intersection
. Let
. Since
, then
. By dropping this altitude, we can also see two similar triangles,
and
. Since
is
, and
, we get that
. Now, if we redraw another diagram just of
, we get that
because of the altitude geometric mean theorem which states that the altitude squared is equal to the product of the two lengths that it divides the base into (for any right triangle). Now expanding, simplifying, and dividing by the GCF, we get
. This factors to
. Since lengths cannot be negative,
. Since
,
. So
~ Solution by Ultraman
~ Diagram by ciceronii
Solution 2 (Coordinates)
Let the points be
,
,
,
,and
, respectively. Since
lies on line
, we know that
. Furthermore, since
,
lies on the circle with diameter
, so
. Solving for
and
with these equations, we get the solutions
and
. We immediately discard the
solution as
should be negative. Thus, we conclude that
.
Solution 3 (Trigonometry)
Let and
Using Law of Sines on
we get
and LoS on
yields
Divide the two to get
Now,
and solve the quadratic, taking the positive solution (C is acute) to get
So if
then
and
By Pythagorean Theorem,
and the answer is
(This solution is incomplete, can someone complete it please-Lingjun) ok Latex edited by kc5170
We could use the famous m-n rule in trigonometry in with Point
[Unable to write it here.Could anybody write the expression]
. We will find that
is an angle bisector of
(because we will get
).
Therefore by converse of angle bisector theorem
. By using Pythagorean theorem, we have values of
and
.
Computing
. Adding the areas of
and
, hence the answer is
.
By: Math-Amaze Latex: Catoptrics.
Solution 4 (Answer Choices)
We know that the big triangle has area 300. Using the answer choices would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. We guess that the legs are equal to and
, and because the hypotenuse is 20 we get
. Testing small numbers, we get that when
and
,
is indeed a square. The area of the triangle is thus
, so the answer is
.
~tigershark22
Solution 5 (LoC)
Denote as
. By the Law of Cosine:
Adding these up yields:
By the quadratic formula,
.
Observe:
.
Thus the desired area is
~qwertysri987
Solution 6 (basic vectors/coordinates)
Let and
. Then
and
lies on the line
So the coordinates of
are
We can make this a vector problem.
We notice that point
forms a right angle, meaning vectors
and
are orthogonal, and their dot-product is
.
We determine and
to be
and
, respectively. (To get this, we use the fact that
and similarly,
)
Equating the cross-product to gets us the quadratic
The solutions are
Since
clearly has a more negative x-coordinate than
, we take
. So
From here, there are multiple ways to get the area of to be
, and since the area of
is
, we get our final answer to be
-PureSwag
Video Solutions
Video Solution 1
On The Spot STEM
https://www.youtube.com/watch?v=hIdNde2Vln4
Video Solution 2
https://www.youtube.com/watch?v=sHrjx968ZaM&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=2 ~ MathEx
Video Solution 3
Education, The Study of Everything
Video Solution 4
The Beauty Of Math https://www.youtube.com/watch?v=RKlG6oZq9so&ab_channel=TheBeautyofMath
Video Solution 5
https://youtu.be/R220vbM_my8?t=658
(amritvignesh0719062.0)
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
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