Difference between revisions of "2018 AMC 10A Problems/Problem 24"
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+ | {{duplicate|[[2018 AMC 10A Problems/Problem 24|2018 AMC 10A #24]] and [[2018 AMC 12A Problems/Problem 18|2018 AMC 12A #18]]}} | ||
+ | |||
== Problem == | == Problem == | ||
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\textbf{(E) }80 \qquad | \textbf{(E) }80 \qquad | ||
</math> | </math> | ||
+ | |||
+ | ==Diagram== | ||
+ | [[File:2018_amc_10a_24_accurate_diagram.png]] | ||
+ | |||
== Solution 1 == | == Solution 1 == | ||
− | Let <math>BC = a</math>, <math>BG = x</math>, <math>GC = y</math>, and the length of the perpendicular | + | Let <math>BC = a</math>, <math>BG = x</math>, <math>GC = y</math>, and the length of the perpendicular from <math>BC</math> through <math>A</math> be <math>h</math>. By angle bisector theorem, we have that <cmath>\frac{50}{x} = \frac{10}{y},</cmath> where <math>y = -x+a</math>. Therefore substituting we have that <math>BG=\frac{5a}{6}</math>. By similar triangles, we have that <math>DF=\frac{5a}{12}</math>, and the height of this trapezoid is <math>\frac{h}{2}</math>. Then, we have that <math>\frac{ah}{2}=120</math>. We wish to compute <math>\frac{5a}{8}\cdot\frac{h}{2}</math>, and we have that it is <math>\boxed{\textbf{(D) }75}</math> by substituting. |
− | |||
== Solution 2 == | == Solution 2 == | ||
− | For this problem, we have <math>\triangle{ADE}\sim\triangle{ABC}</math> because of SAS and <math>DE = \frac{BC}{2}</math>. Therefore, <math>\bigtriangleup ADE</math> is a quarter of the area of <math>\bigtriangleup ABC</math>, which is <math>30</math>. Subsequently, we can compute the area of quadrilateral <math>BDEC</math> to be <math>120 - 30 = 90</math>. Using the angle bisector theorem in the same fashion as the previous problem, we get that <math>\overline{BG}</math> is <math>5</math> times the length of <math>\overline{GC}</math>. We want the larger piece, as described by the problem. Because the heights are identical, one area is <math>5</math> times the other, and <math>\frac{5}{6} \cdot 90 = \boxed{75}</math>. | + | For this problem, we have <math>\triangle{ADE}\sim\triangle{ABC}</math> because of SAS and <math>DE = \frac{BC}{2}</math>. Therefore, <math>\bigtriangleup ADE</math> is a quarter of the area of <math>\bigtriangleup ABC</math>, which is <math>30</math>. Subsequently, we can compute the area of quadrilateral <math>BDEC</math> to be <math>120 - 30 = 90</math>. Using the angle bisector theorem in the same fashion as the previous problem, we get that <math>\overline{BG}</math> is <math>5</math> times the length of <math>\overline{GC}</math>. We want the larger piece, as described by the problem. Because the heights are identical, one area is <math>5</math> times the other, and <math>\frac{5}{6} \cdot 90 = \boxed{\textbf{(D) }75}</math>. |
== Solution 3 == | == Solution 3 == | ||
− | The area of <math>\bigtriangleup ABG</math> to the area of <math>\bigtriangleup ACG</math> is <math>5:1</math> | + | The ratio of the <math>\overline{BG}</math> to <math>\overline{GC}</math> is <math>5:1</math> by the Angle Bisector Theorem, so area of <math>\bigtriangleup ABG</math> to the area of <math>\bigtriangleup ACG</math> is also <math>5:1</math> (They have the same height). Therefore, the area of <math>\bigtriangleup ABG</math> is <math>\frac{5}{5+1}\times120=100</math>. Since <math>\overline{DE}</math> is the midsegment of <math>\bigtriangleup ABC</math>, so <math>\overline{DF}</math> is the midsegment of <math>\bigtriangleup ABG</math> . Thus, the ratio of the area of <math>\bigtriangleup ADF</math> to the area of <math>\bigtriangleup ABG</math> is <math>1:4</math>, so the area of <math>\bigtriangleup ACG</math> is <math>\frac{1}{4}\times100=25</math>. Therefore, the area of quadrilateral <math>FDBG</math> is <math>[ABG]-[ADF]=100-25=\boxed{\textbf{(D) }75}</math> |
− | |||
==Solution 4 == | ==Solution 4 == | ||
− | The area of quadrilateral <math>FDBG</math> is the area of <math>\bigtriangleup ABG</math> minus the area of <math>\bigtriangleup ADF</math>. Notice, <math>\overline{DE} || \overline{BC}</math>, so <math>\bigtriangleup ABG \sim \bigtriangleup ADF</math>, and since <math>\overline{AD}:\overline{AB}=1:2</math>, the area of <math>\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4</math>. Given that the area of <math>\bigtriangleup ABC</math> is <math>120</math>, using <math>\frac{bh}{2}</math> on side <math>AB</math> yields <math>\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}</math>. Using the Angle Bisector Theorem, <math>\overline{BG}:\overline{BC}=50:(10+50)=5:6</math>, so the height of <math>\bigtriangleup ABG: \bigtriangleup ACB=5:6</math>. Therefore our answer is <math>\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}</math> | + | The area of quadrilateral <math>FDBG</math> is the area of <math>\bigtriangleup ABG</math> minus the area of <math>\bigtriangleup ADF</math>. Notice, <math>\overline{DE} || \overline{BC}</math>, so <math>\bigtriangleup ABG \sim \bigtriangleup ADF</math>, and since <math>\overline{AD}:\overline{AB}=1:2</math>, the area of <math>\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4</math>. Given that the area of <math>\bigtriangleup ABC</math> is <math>120</math>, using <math>\frac{bh}{2}</math> on side <math>AB</math> yields <math>\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}</math>. Using the Angle Bisector Theorem, <math>\overline{BG}:\overline{BC}=50:(10+50)=5:6</math>, so the height of <math>\bigtriangleup ABG: \bigtriangleup ACB=5:6</math>. Therefore our answer is <math>\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{\textbf{(D) }75}</math> |
− | -Solution by | + | |
+ | ==Solution 5 (Trigonometry) == | ||
+ | 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. | ||
+ | |||
+ | = | ||
+ | |||
+ | ==Solution 7 (Barycentrics) == | ||
+ | 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{\textbf{(D) }75}.</math> | ||
+ | <math>\sim</math>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== | ==See Also== |
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.