AMC 12A 2024 (Problem 20)ProbabilityHardPoints 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 TopicsCoreGeometric ProbabilityMajorReciprocal FunctionsArea FormulasCheck AnswerYour answer:ABCDECheckHints (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]Related Problems (17)AMC 10B/12B 2024 (Problem 14/9)AMC 10A 2025 (Problem 25)MathCounts 2026 Chapter Sprint Round (Problem 28)AMC 12A 2025 (Problem 10)AMC 10B 2025 (Problem 6)AMC 10B 2025 (Problem 12)AMC 12B 2025 (Problem 21)AMC 8 2026 (Problem 11)AIME II 2026 (Problem 3)2024 AMC 10A Problem 14AMC 10B 2024 (Problem 10)AMC 12B 2024 (Problem 19)AMC 12B 2024 (Problem 20)MATHCOUNTS State Team 2025 (Problem 9)AMC 10A 2025 (Problem 10)2024 AMC 12A Problem 24AMC 12B 2024 (Problem 12)View all related problems →