AMC 12A 2024 (Problem 19)Cyclic quadrilateral ABCDABCDABCD has lengths BC=CD=3BC=CD=3BC=CD=3 and DA=5DA=5DA=5 with ∠CDA=120∘\angle CDA=120^\circ∠CDA=120∘. What is the length of the shorter diagonal of ABCDABCDABCD?(A) 317\text{(A)}\;\frac{31}{7}(A)731(B) 337\text{(B)}\;\frac{33}{7}(B)733(C) 5\text{(C)}\;5(C)5(D) 397\text{(D)}\;\frac{39}{7}(D)739(E) 417\text{(E)}\;\frac{41}{7}(E)741Related TopicsToolkit 19 — Factoring quadratic by groupingToolkit 21 — Quadratic FormulaToolkit 57 — Famous Trigonometric ValuesToolkit 58 — Law of CosinesToolkit 59 — Ptolemy's TheoremToolkit 61 — 4 Rules of Cyclic QuadrilateralsHints (8)Hint 1Find ACACAC.Hint 2By Toolkit 58 — Law of Cosines in △ADC\triangle ADC△ADC:AC2=52+32−2(5)(3)cos120∘AC^2=5^2+3^2-2(5)(3)\cos120^\circAC2=52+32−2(5)(3)cos120∘=25+9−2(5)(3)(−12)=49⇒AC=7=25+9-2(5)(3)\left(-\frac12\right)=49\Rightarrow AC=7=25+9−2(5)(3)(−21)=49⇒AC=7Hint 3Find ∠ABC\angle ABC∠ABC and ABABAB.Hint 4∠ABC=180∘−120∘=60∘\angle ABC=180^\circ-120^\circ=60^\circ∠ABC=180∘−120∘=60∘Hint 5By Toolkit 58 — Law of Cosines in △ABC\triangle ABC△ABC:72=32+AB2−2(3)(AB)cos60∘7^2=3^2+AB^2-2(3)(AB)\cos60^\circ72=32+AB2−2(3)(AB)cos60∘49=9+AB2−2(3)(AB)(12)49=9+AB^2-2(3)(AB)\left(\frac12\right)49=9+AB2−2(3)(AB)(21)0=AB2−3AB−400=AB^2-3AB-400=AB2−3AB−40Hint 6By Toolkit 19 — Factoring quadratic by groupingorToolkit 21 — Quadratic FormulaFind ABABAB.Hint 7AB2−3AB−40=0AB^2-3AB-40=0AB2−3AB−40=0By Toolkit 19 — Factoring quadratic by grouping:(AB−8)(AB+5)=0⇒AB−8=0 or AB+5=0(AB-8)(AB+5)=0\Rightarrow AB-8=0\text{ or }AB+5=0(AB−8)(AB+5)=0⇒AB−8=0 or AB+5=0⇒AB=8 or −5\Rightarrow AB=8\text{ or }-5⇒AB=8 or −5AB>0⇒AB=8AB>0\Rightarrow AB=8AB>0⇒AB=8orToolkit 21 — Quadratic Formula:AB=−(−3)±(−3)2−4(1)(−40)2AB=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-40)}}{2}AB=2−(−3)±(−3)2−4(1)(−40)=3±1692=3±132=\frac{3\pm\sqrt{169}}{2}=\frac{3\pm13}{2}=23±169=23±13=162 or −102=8,−5=\frac{16}{2}\text{ or }\frac{-10}{2}=8,-5=216 or 2−10=8,−5AB>0⇒AB=8AB>0\Rightarrow AB=8AB>0⇒AB=8Hint 8By Toolkit 59 — Ptolemy's Theorem:AB⋅CD+BC⋅AD=BD⋅ACAB\cdot CD+BC\cdot AD=BD\cdot ACAB⋅CD+BC⋅AD=BD⋅AC(8)(3)+(3)(5)=BD(7)(8)(3)+(3)(5)=BD(7)(8)(3)+(3)(5)=BD(7)24+15=7BD24+15=7BD24+15=7BDBD=397BD=\frac{39}{7}BD=739Final Answer(D) 397\frac{39}{7}739
