2019 AMC 8 Problems/Problem 24
Contents
Problem 24
In triangle , point
divides side
s that
. Let
be the midpoint of
and left
be the point of intersection of line
and line
. Given that the area of
is
, what is the area of
?
Solution 1
Draw on
such that
is parallel to
. That makes triangles
and
congruent since
.
so
. Since
(
and
, so
), the altitude of triangle
is equal to
of the altitude of
. The area of
is
, so the area of
~heeeeeeeheeeee
Solution 2 (Mass Points)
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us.
First, we assign a mass of to point
. We figure out that
has a mass of
since
. Then, by adding
, we get that point
has a mass of 3. By equality, point
has a mass of 3 also.
Now, we add for point
and
for point
.
Now, is a common base for triangles
and
, so we figure out that the ratios of the areas is the ratios of the heights which is
. So,
's area is one third the area of
, and we know the area of
is
the area of
since they have the same heights but different bases.
So we get the area of as
-Brudder
Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of
over the product of the mass points of
which is
which also yields
-Brudder
Solution 3
is equal to
. The area of triangle
is equal to
because it is equal to on half of the area of triangle
, which is equal to one third of the area of triangle
, which is
. The area of triangle
is the sum of the areas of triangles
and
, which is respectively
and
. So,
is equal to
=
, so the area of triangle
is
. That minus the area of triangle
is
. ~~SmileKat32
Solution 4 (Similar Triangles)
Extend to
such that
as shown:
Then
and
. Since
, triangle
has four times the area of triangle
. Since
, we get
.
Since is also
, we have
because triangles
and
have the same height and same areas and so their bases must be the congruent. Thus triangle
has twice the side lengths and therefore four times the area of triangle
, giving
.
(Credit to MP8148 for the idea)
Solution 5 (Area Ratios)
As before we figure out the areas labeled in the diagram. Then we note that
Solving gives
.
(Credit to scrabbler94 for the idea)
Solution 6(Coordbash)
Let be a right triangle, and
Let
The line can be described with the equation
The line can be described with
Solving, we get and
Now we can find
-Trex4days
Solution 7
(the median divides the area of the triangle into two equal parts)
Construction: Draw a circumcircle around with
as is diameter. Extend
to
such that it meets the circle at
. Draw line
.
(Since
is cyclic)
But is common in both with an area of 60. So,
.
\therefore (SAS Congruency Theorem).
In , let
be the median of
.
Which means
Rotate to meet
at
and
at
.
will fit exactly in
(both are radii of the circle). From the above solutions,
.
is a radius and
is half of it implies
=
.
Which means
Thus
~phoenixfire & flamewavelight
Solution 8
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 | ||
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