2017 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 19 Medium

Let ABC be an equilateral triangle. Extend side \overline{AB} beyond B to a point {{B}^{\prime }} so that B{{B}^{\prime }}=3AB. Similarly, extend side \overline{BC} beyond C to a point {{C}^{\prime }} so that C{{C}^{\prime }}=3BC, and extend side \overline{CA} beyond A to a point {{A}^{\prime }} so that A{{A}^{\prime }}=3CA. What is the ratio of the area of \triangle {{A}^{\prime }}{{B}^{\prime }}{{C}^{\prime }} to the area of \triangle ABC? (2017 AMC 10B Problem, Question#19)

  • A.

    9:1

  • B.

    16:1

  • C.

    25:1

  • D.

    36:1

  • E.

    37:1

Answer:E

Solution 1

Note that by symmetry, \triangle A'B'C' is also equilateral. Therefore, we only need to find one of the sides of A'B'C' to determine the area ratio. WLOG, let AB=BC=CA=1. Therefore,BB'=3  and BC'=4. Also,\angle B'BC'=120^\circ , so by the Law of Cosines,B'C'=\sqrt{37} . Therefore, the answer is \left( \sqrt{37}\right)^{2}:1^{2}=37:1.

Solution 2

As mentioned in the first solution, \triangle A'B'C' is equilateral. WLOG, let AB=2. Let D be on the line passing through AB such that A'D is perpendicular to AB. Note that \triangle A'DA is a 30-60-90 with right angle at D. Since AA'=6,AD=3 and A'D=3\sqrt 3. So we know that .db'=11 Note that \triangle A'DB' is a right triangle with right angle at D. So by the Pythagorean theorem, we find A'B'=\sqrt{\left( 3\sqrt{3}\right)^{2}+11^{2}}=2\sqrt{37}.Therefore, the answer is \left( 2\sqrt{37}\right)^{2}：2^{2}=37：1.

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