Difference between revisions of "2018 AMC 10A Problems/Problem 24"

Latest revision as of 16:13, 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.

Problem

Triangle $ABC$ with $AB=50$ and $AC=10$ has area $120$ . Let $D$ be the midpoint of $\overline{AB}$ , and let $E$ be the midpoint of $\overline{AC}$ . The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G$ , respectively. What is the area of quadrilateral $FDBG$ ?

$\textbf{(A) }60 \qquad \textbf{(B) }65 \qquad \textbf{(C) }70 \qquad \textbf{(D) }75 \qquad \textbf{(E) }80 \qquad$

Diagram

Solution 1

Let $BC = a$ , $BG = x$ , $GC = y$ , and the length of the perpendicular from $BC$ through $A$ be $h$ . By angle bisector theorem, we have that $\frac{50}{x} = \frac{10}{y},$ where $y = -x+a$ . Therefore substituting we have that $BG=\frac{5a}{6}$ . By similar triangles, we have that $DF=\frac{5a}{12}$ , and the height of this trapezoid is $\frac{h}{2}$ . Then, we have that $\frac{ah}{2}=120$ . We wish to compute $\frac{5a}{8}\cdot\frac{h}{2}$ , and we have that it is $\boxed{\textbf{(D) }75}$ by substituting.

Solution 2

For this problem, we have $\triangle{ADE}\sim\triangle{ABC}$ because of SAS and $DE = \frac{BC}{2}$ . Therefore, $\bigtriangleup ADE$ is a quarter of the area of $\bigtriangleup ABC$ , which is $30$ . Subsequently, we can compute the area of quadrilateral $BDEC$ to be $120 - 30 = 90$ . Using the angle bisector theorem in the same fashion as the previous problem, we get that $\overline{BG}$ is $5$ times the length of $\overline{GC}$ . We want the larger piece, as described by the problem. Because the heights are identical, one area is $5$ times the other, and $\frac{5}{6} \cdot 90 = \boxed{\textbf{(D) }75}$ .

Solution 3

The ratio of the $\overline{BG}$ to $\overline{GC}$ is $5:1$ by the Angle Bisector Theorem, so area of $\bigtriangleup ABG$ to the area of $\bigtriangleup ACG$ is also $5:1$ (They have the same height). Therefore, the area of $\bigtriangleup ABG$ is $\frac{5}{5+1}\times120=100$ . Since $\overline{DE}$ is the midsegment of $\bigtriangleup ABC$ , so $\overline{DF}$ is the midsegment of $\bigtriangleup ABG$ . Thus, the ratio of the area of $\bigtriangleup ADF$ to the area of $\bigtriangleup ABG$ is $1:4$ , so the area of $\bigtriangleup ACG$ is $\frac{1}{4}\times100=25$ . Therefore, the area of quadrilateral $FDBG$ is $[ABG]-[ADF]=100-25=\boxed{\textbf{(D) }75}$

Solution 4

The area of quadrilateral $FDBG$ is the area of $\bigtriangleup ABG$ minus the area of $\bigtriangleup ADF$ . Notice, $\overline{DE} || \overline{BC}$ , so $\bigtriangleup ABG \sim \bigtriangleup ADF$ , and since $\overline{AD}:\overline{AB}=1:2$ , the area of $\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4$ . Given that the area of $\bigtriangleup ABC$ is $120$ , using $\frac{bh}{2}$ on side $AB$ yields $\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}$ . Using the Angle Bisector Theorem, $\overline{BG}:\overline{BC}=50:(10+50)=5:6$ , so the height of $\bigtriangleup ABG: \bigtriangleup ACB=5:6$ . Therefore our answer is $\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{\textbf{(D) }75}$

Solution 5 (Trigonometry)

We try to find the area of quadrilateral $FDBG$ by subtracting the area outside the quadrilateral but inside triangle $ABC$ . Note that the area of $\triangle ADE$ is equal to $\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}$ and the area of triangle $ABC$ is equal to $\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A$ . The ratio $\frac{[ADE]}{[ABC]}$ is thus equal to $\frac{1}{4}$ and the area of triangle $ADE$ is $\frac{1}{4} \cdot 120 = 30$ . Let side $BC$ be equal to $6x$ , then $BG = 5x, GC = x$ by the angle bisector theorem. Similarly, we find the area of triangle $AGC$ to be $\frac{1}{2} \cdot 10 \cdot x \cdot \sin C$ and the area of triangle $ABC$ to be $\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C$ . A ratio between these two triangles yields $\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}$ , so $[AGC] = 20$ . Now we just need to find the area of triangle $AFE$ and subtract it from the combined areas of $[ADE]$ and $[ACG]$ , since we count it twice. Note that the angle bisector theorem also applies for $\triangle ADE$ and $\frac{AE}{AD} = \frac{1}{5}$ , so thus $\frac{EF}{ED} = \frac{1}{6}$ and we find $[AFE] = \frac{1}{6} \cdot 30 = 5$ , and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$ , and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \boxed{\textbf{(D) }75}$ , and we are done.

=

Solution 7 (Barycentrics)

Let our reference triangle be $\triangle ABC$ . Consequently, we have $A=(1,0,0)$ , $B=(0,1,0)$ , $C=(0,0,1).$ Since $D$ is the midpoint of $\overline{AB}$ , we have that $D=(1:1:0)$ . Similarly, we have $E=(1:0:1).$ Hence, the line through $D$ and $E$ is given by the equation

$0 = \begin{vmatrix} x & y & z\\ 1 & 1 & 0\\ 1 & 0 & 1 \end{vmatrix}$

Additionally, since all points on $\overline{AG}$ are characterized by $(t:1:5)$ , we may plug in for $x,y,z$ to get $t=6$ . Thus, we have $F=(6:1:5).$ Now, we homogenize the coordinates for $F D, B, G$ to get $F=(\frac{1}{2}, \frac{5}{12}, \frac{1}{12})$ , $D=(\frac{1}{2}, \frac{1}{2}, 0)$ , $B=(0,1,0)$ , $G=(0, \frac{1}{6}, \frac{5}{6})$

Splitting $[FBGD]$ into $[ DBG ] + [ FDG],$ we may now evaluate the two determinants:

$\begin{vmatrix} \frac{1}{2} & \frac{1}{2} & 0\\ 0 & 1 & 0\\ 0 & \frac{1}{6} & \frac{5}{6} \end{vmatrix}$ $\begin{vmatrix} \frac{1}{2} & \frac{1}{12} & \frac{5}{12}\\ \frac{1}{2} & \frac{1}{2} & 0\\ 0 & \frac{5}{6} & \frac{1}{6} \end{vmatrix}.$

After simplification, we get $\frac{5}{12}$ and $\frac{5}{24}$ , respectively. Summing, we get $\frac{15}{24}.$ Hence, $[FBGD]=\frac{15}{24} \cdot 120 = \fbox{\textbf{(D) }75}.$ $\sim$ 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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2018 AMC 10A Problems/Problem 24|2018 AMC 10A #24]] and [[2018 AMC 12A Problems/Problem 18|2018 AMC 12A #18]]}}
 == Problem ==
@@ Line 10: / Line 12: @@
 \textbf{(E) }80 \qquad
 </math>
+==Diagram==
+[[File:2018_amc_10a_24_accurate_diagram.png]]
 == Solution 1 ==
-Let <math>BC = a</math>, <math>BG = x</math>, <math>GC = y</math>, and the length of the perpendicular to <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{75}</math> by substituting.
+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.
-(rachanamadhu)
 ==  Solution 2 ==
-<math>\overline{DE}</math> is midway from <math>A</math> to <math>\overline{BC}</math>, 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 - 90 = 30</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 ==
-The area of <math>\bigtriangleup ABG</math> to the area of <math>\bigtriangleup ACG</math> is <math>5:1</math> by Law of Sines. So the area of <math>\bigtriangleup ABG</math> is <math>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> . So the area of <math>\bigtriangleup ACG</math>  to the area of <math>\bigtriangleup ABG</math> is <math>1:4</math> , so the area of <math>\bigtriangleup ACG</math> is <math>25</math>, by similar triangles. Therefore the area of quad <math>FDBG</math> is <math>100-25=\boxed{75}</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>
-(steakfails)
 ==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>, and so 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 ktong
+==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>
+=
+\begin{vmatrix}
+x & y & z\\
+& 1 & 0\\
+& 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\\
+& 1 & 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\\
+& \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==
@@ Line 32: / Line 85: @@
 {{AMC12 box|year=2018|ab=A|num-b=17|num-a=19}}
 {{MAA Notice}}
+[[Category:Intermediate Geometry Problems]]

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AMC 10 Problems and Solutions

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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