AMC 12A 2025 (Problem 11)The orthocenter of a triangle is the concurrent intersection of the three (possibly extended) altitudes. What is the sum of the coordinates of the orthocenter of the triangle whose vertices are A(2,31)A(2,31)A(2,31), B(8,27)B(8,27)B(8,27), and C(18,27)C(18,27)C(18,27)?(A) 5\text{(A)}\;5(A)5(B) 17\text{(B)}\;17(B)17(C) 10+417+213\text{(C)}\;10+4\sqrt{17}+2\sqrt{13}(C)10+417+213(D) 1133\text{(D)}\;\frac{113}{3}(D)3113(E) 54\text{(E)}\;54(E)54Related TopicsCoreToolkit 95 — Coordinate GeometryHints (8)Hint 1If you also draw ABCABCABC as an acute triangle, the calculation would remain the same. So, there is no need to draw an accurate figure for this problem.Hint 2H=AE∩BDH=AE\cap BDH=AE∩BDHint 3Find sBDs_{BD}sBD.Hint 4Slope of ACACAC:sAC=31−272−18=4−16=−14s_{AC}=\frac{31-27}{2-18}=\frac{4}{-16}=-\frac14sAC=2−1831−27=−164=−41Since AC⊥BDAC\perp BDAC⊥BD,By Toolkit 95 — Coordinate Geometry:sBD=−1sAC=−1−14=4s_{BD}=-\frac{1}{s_{AC}}=-\frac{1}{-\frac14}=4sBD=−sAC1=−−411=4Hint 5BD:y−27=4(x−8)BD:\quad y-27=4(x-8)BD:y−27=4(x−8)Hint 6Find the equation of AEAEAE.Hint 7sBC=0s_{BC}=0sBC=0BCBCBC is horizontal ⇒AE\Rightarrow AE⇒AE is vertical ⇒\Rightarrow⇒AE:x=2AE:\quad x=2AE:x=2Hint 8From Hints 5 and 7:y−27=4(2−8)=−24y-27=4(2-8)=-24y−27=4(2−8)=−24⟹y=3\Longrightarrow y=3⟹y=3⟹H(2,3)\Longrightarrow H(2,3)⟹H(2,3)⟹Ans=2+3=5\Longrightarrow \text{Ans}=2+3=5⟹Ans=2+3=5Final Answer(A) 555Related Problems (2)AMC 10A 2025 (Problem 20)AMC 10A 2025 (Problem 22)
