AMC 10A/12A 2022 (Problems 23/20)Isosceles trapezoid ABCDABCDABCD has parallel sides AD‾\overline{AD}AD and BC‾\overline{BC}BC, with BC<ADBC<ADBC<AD and AB=CDAB=CDAB=CD. There is a point PPP in the plane such that PA=1PA=1PA=1, PB=2PB=2PB=2, PC=3PC=3PC=3, and PD=4PD=4PD=4. What is BCAD\frac{BC}{AD}ADBC?(A) 14\text{(A)}\;\frac14(A)41(B) 13\text{(B)}\;\frac13(B)31(C) 12\text{(C)}\;\frac12(C)21(D) 23\text{(D)}\;\frac23(D)32(E) 34\text{(E)}\;\frac34(E)43Related TopicsCoreDual Pythagorean TheoremTrapezoid Strategy: Draw the AltitudesMinorAlgebraic IdentitiesCheck AnswerYour answer:ABCDECheckHints (8)Hint 1Draw the figure.Hint 2By Trapezoid Strategy: Draw the Altitudes, draw the altitudes and use Dual Pythagorean Theorem.Hint 342−12=PD2−PA2=(x+z+x+y)2−y24^2-1^2=PD^2-PA^2=(x+z+x+y)^2-y^242−12=PD2−PA2=(x+z+x+y)2−y2⟹(2x+y+z)2−y2=15\Longrightarrow (2x+y+z)^2-y^2=15⟹(2x+y+z)2−y2=1532−22=PC2−PB2=(x+z)2−x23^2-2^2=PC^2-PB^2=(x+z)^2-x^232−22=PC2−PB2=(x+z)2−x2⟹(x+z)2−x2=5\Longrightarrow (x+z)^2-x^2=5⟹(x+z)2−x2=5Hint 4BCAD=zy+x+z+x+y=z2x+2y+z=12(x+yz)+1\frac{BC}{AD}=\frac{z}{y+x+z+x+y}=\frac{z}{2x+2y+z}=\frac{1}{2\left(\frac{x+y}{z}\right)+1}ADBC=y+x+z+x+yz=2x+2y+zz=2(zx+y)+11Hint 5By Hint 3,{(2x+y+z)2−y2=15(x+z)2−x2=5\begin{cases}(2x+y+z)^2-y^2=15\\[4pt](x+z)^2-x^2=5\end{cases}{(2x+y+z)2−y2=15(x+z)2−x2=5⟹(2x+y+z)2−(x+z)2+x2−y2=10\Longrightarrow (2x+y+z)^2-(x+z)^2+x^2-y^2=10⟹(2x+y+z)2−(x+z)2+x2−y2=10⟹(x+y)(3x+y+2z)+(x−y)(x+y)=10\Longrightarrow (x+y)(3x+y+2z)+(x-y)(x+y)=10⟹(x+y)(3x+y+2z)+(x−y)(x+y)=10⟹(x+y)(4x+2z)=10\Longrightarrow (x+y)(4x+2z)=10⟹(x+y)(4x+2z)=10⟹(x+y)(2x+z)=5\Longrightarrow (x+y)(2x+z)=5⟹(x+y)(2x+z)=5Hint 6By Hint 3,(x+z)2−x2=5(x+z)^2-x^2=5(x+z)2−x2=5⟹(x+z+x)(x+z−x)=5\Longrightarrow (x+z+x)(x+z-x)=5⟹(x+z+x)(x+z−x)=5⟹(2x+z)z=5\Longrightarrow (2x+z)z=5⟹(2x+z)z=5Hint 7By Hints 5 and 6,{(x+y)(2x+z)=5(2x+z)z=5\begin{cases}(x+y)(2x+z)=5\\[4pt](2x+z)z=5\end{cases}{(x+y)(2x+z)=5(2x+z)z=5⟹x+y=z\Longrightarrow x+y=z⟹x+y=zHint 8By Hints 4 and 7,BCAD=12(x+yz)+1=13\frac{BC}{AD}=\frac{1}{2\left(\frac{x+y}{z}\right)+1}=\frac13ADBC=2(zx+y)+11=31Final Answer(B) 13\frac1331Related Problems (4)AMC 10A/12A 2025 (Problem 23/16)AIME I 2025 (Problem 6)AMC 8 2025 (Problem 24)AMC 12A 2025 (Problem 20)
