Difference between revisions of "2018 AMC 10A Problems/Problem 24"
Pi is 3.14 (talk | contribs) (→Video Solution) |
|||
Line 66: | Line 66: | ||
Now we can solve for X by adding up all the sums. X + 3X + 5X + 15X = 120, so X = 5. Since we want to find <math>GFDB</math>, we substitute 5 for 15X to get <math>\boxed{75}</math>. | Now we can solve for X by adding up all the sums. X + 3X + 5X + 15X = 120, so X = 5. Since we want to find <math>GFDB</math>, we substitute 5 for 15X to get <math>\boxed{75}</math>. | ||
<math>\sim</math>krishkhushi09 | <math>\sim</math>krishkhushi09 | ||
+ | |||
+ | ==Solution 7: Barycentrics (overkill) == | ||
+ | Let our reference triangle be <math>\triangle ABC</math>. Consequently, we have <math>A=(1,0,0)</math>, <math>B=(0,1,0)</math>, <math>C=(0,0,1).</math> Since <math>D</math> is the midpoint of <math>\overline{AB}</math>, we have that <math>D=(1:1:0)</math>. Similarly, we have <math>E=(1:0:1).</math> Hence, the line through <math>D</math> and <math>E</math> is given by the equation | ||
+ | |||
+ | <cmath> | ||
+ | 0 = | ||
+ | \begin{vmatrix} | ||
+ | x & y & z\\ | ||
+ | 1 & 1 & 0\\ | ||
+ | 1 & 0 & 1 | ||
+ | \end{vmatrix} | ||
+ | </cmath> | ||
+ | |||
+ | Additionally, since all points on <math>\overline{AG}</math> are characterized by <math>(t:1:5)</math>, we may plug in for <math>x,y,z</math> to get <math>t=6</math>. Thus, we have <math>F=(6:1:5).</math> Now, we homogenize the coordinates for <math>F D, B, G</math> to get <math>F=(\frac{1}{2}, \frac{5}{12}, \frac{1}{12})</math>, <math>D=(\frac{1}{2}, \frac{1}{2}, 0)</math>, <math>B=(0,1,0)</math>, <math>G=(0, \frac{1}{6}, \frac{5}{6})</math> | ||
+ | |||
+ | |||
+ | |||
+ | Splitting <math>[FBGD]</math> into <math>[ DBG ] + [ FDG],</math> we may now evaluate the two determinants: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{vmatrix} | ||
+ | \frac{1}{2} & \frac{1}{2} & 0\\ | ||
+ | 0 & 1 & 0\\ | ||
+ | 0 & \frac{1}{6} & \frac{5}{6} | ||
+ | \end{vmatrix} | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | \begin{vmatrix} | ||
+ | \frac{1}{2} & \frac{1}{12} & \frac{5}{12}\\ | ||
+ | \frac{1}{2} & \frac{1}{2} & 0\\ | ||
+ | 0 & \frac{5}{6} & \frac{1}{6} | ||
+ | \end{vmatrix}. | ||
+ | </cmath> | ||
+ | |||
+ | After simplification, we get <math>\frac{5}{12}</math> and <math>\frac{5}{24}</math>, respectively. Summing, we get <math>\frac{15}{24}.</math> Hence, <math>[FBGD]=\frac{15}{24} \cdot 120 = \fbox{75}.</math> | ||
+ | <math>\sim</math>Math0323 | ||
+ | |||
== Video Solution by Richard Rusczyk == | == Video Solution by Richard Rusczyk == |
Revision as of 15:04, 16 June 2023
- 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
?
Solution 1
Let ,
,
, and the length of the perpendicular to
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: Trig
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 6: Areas
Give triangle
area X. Then, by similarity, since
,
has area 4X. Thus,
has area 3X.
Doing the same for triangle
, we get that triangle
has area Y and quadrilateral
has area 3Y. Since
has the same height as
, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem,
. So,
. Since
has area X, we can write the equation 5X = Y and substitute 5X for Y.
Now we can solve for X by adding up all the sums. X + 3X + 5X + 15X = 120, so X = 5. Since we want to find
, we substitute 5 for 15X to get
.
krishkhushi09
Solution 7: Barycentrics (overkill)
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.