AMC 12B 2025 (Problem 21)Two non-congruent triangles have the same area. Each triangle has sides of length 888 and 999, and the third side of each triangle has integer length. What is the sum of the lengths of the third sides?(A) 20\text{(A)}\;20(A)20(B) 22\text{(B)}\;22(B)22(C) 24\text{(C)}\;24(C)24(D) 26\text{(D)}\;26(D)26(E) 28\text{(E)}\;28(E)28Related TopicsCoreToolkit 54 — Area Formulas and Important Geometry FormulasMajorToolkit 58 — Law of CosinesMinorToolkit 36 — General Solutions of Trigonometric EquationsToolkit 37 — Trigonometric TransformationsToolkit 88 — Triangle InequalityCheck AnswerYour answer:ABCDECheckHints (3)Hint 1[ABC]=12⋅8⋅9sinα[ABC]=\frac12\cdot8\cdot9\sin\alpha[ABC]=21⋅8⋅9sinα[A′B′C′]=12⋅8⋅9sinβ[A'B'C']=\frac12\cdot8\cdot9\sin\beta[A′B′C′]=21⋅8⋅9sinβ[ABC]=[A′B′C′][ABC]=[A'B'C'][ABC]=[A′B′C′]⇒sinα=sinβ\Rightarrow\sin\alpha=\sin\beta⇒sinα=sinβ△ABC≇△A′B′C′\triangle ABC\not\cong\triangle A'B'C'△ABC≅△A′B′C′⇒β=180∘−α\Rightarrow\beta=180^\circ-\alpha⇒β=180∘−αHint 2a2=82+92−2⋅8⋅9cosαa^2=8^2+9^2-2\cdot8\cdot9\cos\alphaa2=82+92−2⋅8⋅9cosαa′2=82+92−2⋅8⋅9cos(180∘−α)a'^2=8^2+9^2-2\cdot8\cdot9\cos(180^\circ-\alpha)a′2=82+92−2⋅8⋅9cos(180∘−α)=82+92+2⋅8⋅9cosα=8^2+9^2+2\cdot8\cdot9\cos\alpha=82+92+2⋅8⋅9cosα⇒a2+a′2=2(82+92)=2(64+81)=290\Rightarrow a^2+a'^2=2(8^2+9^2)=2(64+81)=290⇒a2+a′2=2(82+92)=2(64+81)=290Hint 3WLOG, a≤a′a\le a'a≤a′.a2≤145⇒a≤12a^2\le145\Rightarrow a\le12a2≤145⇒a≤12By Toolkit 88 — Triangle Inequality,a+8>9⇒a>1⇒a≥2a+8>9\Rightarrow a>1\Rightarrow a\ge2a+8>9⇒a>1⇒a≥2a′2=290−a2a'^2=290-a^2a′2=290−a2=290−4, 290−9, 290−16, 290−25, 290−36, 290−49,=290-4,\ 290-9,\ 290-16,\ 290-25,\ 290-36,\ 290-49,=290−4, 290−9, 290−16, 290−25, 290−36, 290−49,290−64, 290−81, 290−100, 290−121, 290−144290-64,\ 290-81,\ 290-100,\ 290-121,\ 290-144290−64, 290−81, 290−100, 290−121, 290−144⇒a′2=286281274265254241226209190169146×××××××××✓×\Rightarrow\begin{array}{rccccccccccc}a'^2=&286&281&274&265&254&241&226&209&190&169&146\\[4pt]&\times&\times&\times&\times&\times&\times&\times&\times&\times&\checkmark&\times\end{array}⇒a′2=286×281×274×265×254×241×226×209×190×169✓146×⇒a′=13, a=11\Rightarrow a'=13,\ a=11⇒a′=13, a=11Ans=a+a′=11+13=24\text{Ans}=a+a'=11+13=24Ans=a+a′=11+13=24Final Answer(C) 242424Related Problems (15)MathCounts 2026 Chapter Sprint Round (Problem 28)AMC 12A 2025 (Problem 10)AMC 10B 2025 (Problem 6)AMC 10B 2025 (Problem 12)AMC 8 2026 (Problem 11)AMC 10B/12B 2024 (Problem 14/9)AMC 12A 2024 (Problem 20)AMC 12A 2024 (Problem 19)View all related problems →
