Difference between revisions of "2018 AMC 10A Problems/Problem 24"
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\textbf{(E) }80 \qquad | \textbf{(E) }80 \qquad | ||
</math> | </math> | ||
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+ | ==Diagram== | ||
+ | [[File:2018_amc_10a_24_accurate_diagram.png]] | ||
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== Solution 1 == | == Solution 1 == | ||
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We try to find the area of quadrilateral <math>FDBG</math> by subtracting the area outside the quadrilateral but inside triangle <math>ABC</math>. Note that the area of <math>\triangle ADE</math> is equal to <math>\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}</math> and the area of triangle <math>ABC</math> is equal to <math>\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A</math>. The ratio <math>\frac{[ADE]}{[ABC]}</math> is thus equal to <math>\frac{1}{4}</math> and the area of triangle <math>ADE</math> is <math>\frac{1}{4} \cdot 120 = 30</math>. Let side <math>BC</math> be equal to <math>6x</math>, then <math>BG = 5x, GC = x</math> by the angle bisector theorem. Similarly, we find the area of triangle <math>AGC</math> to be <math>\frac{1}{2} \cdot 10 \cdot x \cdot \sin C</math> and the area of triangle <math>ABC</math> to be <math>\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C</math>. A ratio between these two triangles yields <math>\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}</math>, so <math>[AGC] = 20</math>. Now we just need to find the area of triangle <math>AFE</math> and subtract it from the combined areas of <math>[ADE]</math> and <math>[ACG]</math>, since we count it twice. Note that the angle bisector theorem also applies for <math>\triangle ADE</math> and <math>\frac{AE}{AD} = \frac{1}{5}</math>, so thus <math>\frac{EF}{ED} = \frac{1}{6}</math> and we find <math>[AFE] = \frac{1}{6} \cdot 30 = 5</math>, and the area outside <math>FDBG</math> must be <math> [ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45</math>, and we finally find <math>[FDBG] = [ABC] - 45 = 120 -45 = \boxed{\textbf{(D) }75}</math>, and we are done. | We try to find the area of quadrilateral <math>FDBG</math> by subtracting the area outside the quadrilateral but inside triangle <math>ABC</math>. Note that the area of <math>\triangle ADE</math> is equal to <math>\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}</math> and the area of triangle <math>ABC</math> is equal to <math>\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A</math>. The ratio <math>\frac{[ADE]}{[ABC]}</math> is thus equal to <math>\frac{1}{4}</math> and the area of triangle <math>ADE</math> is <math>\frac{1}{4} \cdot 120 = 30</math>. Let side <math>BC</math> be equal to <math>6x</math>, then <math>BG = 5x, GC = x</math> by the angle bisector theorem. Similarly, we find the area of triangle <math>AGC</math> to be <math>\frac{1}{2} \cdot 10 \cdot x \cdot \sin C</math> and the area of triangle <math>ABC</math> to be <math>\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C</math>. A ratio between these two triangles yields <math>\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}</math>, so <math>[AGC] = 20</math>. Now we just need to find the area of triangle <math>AFE</math> and subtract it from the combined areas of <math>[ADE]</math> and <math>[ACG]</math>, since we count it twice. Note that the angle bisector theorem also applies for <math>\triangle ADE</math> and <math>\frac{AE}{AD} = \frac{1}{5}</math>, so thus <math>\frac{EF}{ED} = \frac{1}{6}</math> and we find <math>[AFE] = \frac{1}{6} \cdot 30 = 5</math>, and the area outside <math>FDBG</math> must be <math> [ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45</math>, and we finally find <math>[FDBG] = [ABC] - 45 = 120 -45 = \boxed{\textbf{(D) }75}</math>, and we are done. | ||
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==Solution 7 (Barycentrics) == | ==Solution 7 (Barycentrics) == |
Latest revision as of 17:13, 3 March 2024
- The following problem is from both the 2018 AMC 10A #24 and 2018 AMC 12A #18, so both problems redirect to this page.
Contents
Problem
Triangle with
and
has area
. Let
be the midpoint of
, and let
be the midpoint of
. The angle bisector of
intersects
and
at
and
, respectively. What is the area of quadrilateral
?
Diagram
Solution 1
Let ,
,
, and the length of the perpendicular from
through
be
. By angle bisector theorem, we have that
where
. Therefore substituting we have that
. By similar triangles, we have that
, and the height of this trapezoid is
. Then, we have that
. We wish to compute
, and we have that it is
by substituting.
Solution 2
For this problem, we have because of SAS and
. Therefore,
is a quarter of the area of
, which is
. Subsequently, we can compute the area of quadrilateral
to be
. Using the angle bisector theorem in the same fashion as the previous problem, we get that
is
times the length of
. We want the larger piece, as described by the problem. Because the heights are identical, one area is
times the other, and
.
Solution 3
The ratio of the to
is
by the Angle Bisector Theorem, so area of
to the area of
is also
(They have the same height). Therefore, the area of
is
. Since
is the midsegment of
, so
is the midsegment of
. Thus, the ratio of the area of
to the area of
is
, so the area of
is
. Therefore, the area of quadrilateral
is
Solution 4
The area of quadrilateral is the area of
minus the area of
. Notice,
, so
, and since
, the area of
. Given that the area of
is
, using
on side
yields
. Using the Angle Bisector Theorem,
, so the height of
. Therefore our answer is
Solution 5 (Trigonometry)
We try to find the area of quadrilateral by subtracting the area outside the quadrilateral but inside triangle
. Note that the area of
is equal to
and the area of triangle
is equal to
. The ratio
is thus equal to
and the area of triangle
is
. Let side
be equal to
, then
by the angle bisector theorem. Similarly, we find the area of triangle
to be
and the area of triangle
to be
. A ratio between these two triangles yields
, so
. Now we just need to find the area of triangle
and subtract it from the combined areas of
and
, since we count it twice. Note that the angle bisector theorem also applies for
and
, so thus
and we find
, and the area outside
must be
, and we finally find
, and we are done.
=
Solution 7 (Barycentrics)
Let our reference triangle be . Consequently, we have
,
,
Since
is the midpoint of
, we have that
. Similarly, we have
Hence, the line through
and
is given by the equation
Additionally, since all points on are characterized by
, we may plug in for
to get
. Thus, we have
Now, we homogenize the coordinates for
to get
,
,
,
Splitting into
we may now evaluate the two determinants:
After simplification, we get and
, respectively. Summing, we get
Hence,
Math0323
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2018amc10a/469
~ dolphin7
Video Solution by OmegaLearn
https://youtu.be/4_x1sgcQCp4?t=4898
~ pi_is_3.14
See Also
2018 AMC 10A (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 | ||
All AMC 10 Problems and Solutions |
2018 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.