AMC 10B/12B 2025 (Problem 13/10)The altitude to the hypotenuse of a 303030-606060-90∘90^\circ90∘ right triangle is divided into two segments of lengths x<yx<yx<y by the median to the shortest side of the triangle. What is the ratio xx+y\frac{x}{x+y}x+yx?(A) 37\text{(A)}\;\frac{3}{7}(A)73(B) 34\text{(B)}\;\frac{\sqrt3}{4}(B)43(C) 49\text{(C)}\;\frac{4}{9}(C)94(D) 511\text{(D)}\;\frac{5}{11}(D)115(E) 4315\text{(E)}\;\frac{4\sqrt3}{15}(E)1543Related TopicsCoreToolkit 108 — Ratio LemmaMinorToolkit 80 — Trigonometric Ratios in a Right TriangleToolkit 57 — Famous Trigonometric ValuesToolkit 96 — Ratio ManipulationHints (4)Hint 1Hint 2By Toolkit 108 — Ratio Lemma in △ABC\triangle ABC△ABC and line CMCMCM:sinC1sinC2=BMMA⋅ACBC=ACBC=sin60∘=32\frac{\sin C_1}{\sin C_2}=\frac{BM}{MA}\cdot\frac{AC}{BC}=\frac{AC}{BC}=\sin60^\circ=\frac{\sqrt3}{2}sinC2sinC1=MABM⋅BCAC=BCAC=sin60∘=23Hint 3By Toolkit 108 — Ratio Lemma in △CDA\triangle CDA△CDA and line CPCPCP:sinC1sinC2=xy⋅ACCD=xy⋅1sin60∘=xy⋅23\frac{\sin C_1}{\sin C_2}=\frac{x}{y}\cdot\frac{AC}{CD}=\frac{x}{y}\cdot\frac{1}{\sin60^\circ}=\frac{x}{y}\cdot\frac{2}{\sqrt3}sinC2sinC1=yx⋅CDAC=yx⋅sin60∘1=yx⋅32Hint 4By Hints 2 and 3:32=xy⋅23\frac{\sqrt3}{2}=\frac{x}{y}\cdot\frac{2}{\sqrt3}23=yx⋅32⇒xy=34\Rightarrow\frac{x}{y}=\frac34⇒yx=43⇒xx+y=33+4=37\Rightarrow\frac{x}{x+y}=\frac{3}{3+4}=\frac37⇒x+yx=3+43=73Final Answer(A) 37\frac{3}{7}73
