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

Revision as of 09:58, 17 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.

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 $FDBG$ ?

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

Solution 1

Let $BC = a$ , $BG = x$ , $GC = y$ , and the length of the perpendicular to $BC$ 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 $\boxed{75}$ 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 $\frac{5}{6} \cdot 90 = \boxed{75}$ .

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 $[ABG]-[ADF]=100-25=\boxed{75}$

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 $\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}$

Solution 5: Trig

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 $[FDBG] = [ABC] - 45 = 120 -45 = \boxed{75}$ , and we are done.

Solution 6: Areas

$[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label("A",(0,0),SW); label("B",(5,0),SE); label("C",(1,3),N); label("G",(2,2.25),NE); label("D",(2.5,0),S); label("E",(0.5,1.5),NW); label("3Y",(2.5,0.75),N); label("Y",(1,0.2),N); label("X",(0.5,0.5),N); label("3X",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\frac{AC}{AE} = \frac{2}{1}$ , $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$ , we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$ , the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\frac{CG}{GB} = \frac{1}{5}$ . So, $\frac{[AEF]}{[AFD]} = \frac{1}{5}$ . Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label("A",(0,0),SW); label("B",(5,0),SE); label("C",(1,3),N); label("G",(2,2.25),NE); label("D",(2.5,0),S); label("E",(0.5,1.5),NW); label("",(2.5,0.75),N); label("",(1,0.2),N); label("F", (1, 1.5), N); label("",(0.5,1.5),N); label("",(1.25,1.75),N); [/asy]$ 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 $GFDB$ , we substitute 5 for 15X to get $\boxed{75}$ . $\sim$ krishkhushi09

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, $[FBGD]=\frac{15}{24} \cdot 120 = \fbox{75}.$ $\sim$ 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 67: / Line 67: @@
 <math>\sim</math>krishkhushi09
-==Solution 7: Barycentrics ==
+==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

2018 AMC 10A (Problems • Answer Key • Resources)
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