AMC 12A 2024 (Problem 20)Points PPP and QQQ are chosen uniformly and independently at random on sides AB‾\overline{AB}AB and AC‾\overline{AC}AC, respectively, of equilateral triangle △ABC\triangle ABC△ABC. Which of the following intervals contains the probability that the area of △APQ\triangle APQ△APQ is less than half the area of △ABC\triangle ABC△ABC?(A) [38,12]\text{(A)}\;\left[\frac{3}{8},\frac{1}{2}\right](A)[83,21](B) (12,23]\text{(B)}\;\left(\frac{1}{2},\frac{2}{3}\right](B)(21,32](C) (23,34]\text{(C)}\;\left(\frac{2}{3},\frac{3}{4}\right](C)(32,43](D) (34,78]\text{(D)}\;\left(\frac{3}{4},\frac{7}{8}\right](D)(43,87](E) (78,1]\text{(E)}\;\left(\frac{7}{8},1\right](E)(87,1]Related TopicsToolkit 53 — Graphing Functions: Transformations and Absolute Value FunctionsToolkit 54 — Area Formulas and Important Geometry FormulasToolkit 70 — Geometric Approach in ProbabilityHints (9)Hint 1WLOG, assume the side length of the equilateral triangle is 111.[APQ][ABC]=12xysinA12(1)(1)sinA=xy\frac{[APQ]}{[ABC]}=\frac{\frac12xy\sin A}{\frac12(1)(1)\sin A}=xy[ABC][APQ]=21(1)(1)sinA21xysinA=xyHint 20≤x,y≤1Pr(xy<12)=?0\le x,y\le 1 \qquad \Pr\left(xy<\frac12\right)=?0≤x,y≤1Pr(xy<21)=?Hint 3xy<12⇒y<12xxy<\frac12 \Rightarrow y<\frac{1}{2x}xy<21⇒y<2x1Hint 4Graph y=12xy=\frac{1}{2x}y=2x1.Hint 5Probability=shaded region1×1\text{Probability}=\frac{\text{shaded region}}{1\times1}Probability=1×1shaded regionHint 6Calculate an upper bound for the shaded region.Hint 7Probability<[ABCDO]=1−12×122=1−18=78\text{Probability}<[ABCDO]=1-\frac{\frac12\times\frac12}{2}=1-\frac18=\frac78Probability<[ABCDO]=1−221×21=1−81=87Hint 8Calculate a lower bound for the shaded region.Hint 9Probability>[ABECDO]=1−12×12=34\text{Probability}>[ABECDO]=1-\frac12\times\frac12=\frac34Probability>[ABECDO]=1−21×21=43Final Answer(D) (34,78]\left(\frac34,\frac78\right](43,87]
