2018 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 24 Hard

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? (2018 AMC 10A Problem, Question#24)

  • A.

    60

  • B.

    65

  • C.

    70

  • D.

    75

  • E.

    80

Answer:D

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.

For this problem, we have \triangle ADE\backsim \triangle ABC because of \rmSAS and DE=\frac{BC}{2}.

Therefore, \triangle ADE is a quarter of the area of \triangle 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}.

The area of \triangle ABG to the area of \triangle ACG is 5:1 by Law of Sines. So the area of \triangle ABG is 100 .

Since \overline{DE} is the midsegment of \triangle ABC, so \overline{DF} is the midsegment of \triangle ABG. So the area of \triangle ACG to the area of \triangle ABG is 1:4, so the area of \triangle ACG is 25, by similar triangles.

Therefore the area of quad FDBG is 100-25=\boxed{75}.

The area of quadrilateral FDBG is the area of \triangle ABG minus the area of \triangle ADF.

Notice, \overline{DE}//\overline{BC}, so \triangle ABG\backsim \triangle ADF, and since \overline{AD}:\overline{AB}=1:2, the area of \triangle ADF:\triangle ABG=(1:2)^2=1:4. Given that the area of \triangle ABC is 120, using \frac{bh}{2} on side AB yields \dfrac{50h}{2}=120\Rightarrow h=\dfrac{240}{50}=\dfrac{24}{5}. Using the Angle Bisector Theorem, \overline{BG}:\overline{BC}=50:(10+50)=5:6, so the height of \triangle ABG:\triangle ACB=5:6.

Therefore our answer is [FDBG]=[ABG]-[ADF]=[ABG]\left(1-\frac{1}{4}\right)=\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}.

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