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Angles and Triangles | AIME I, 2012 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Angles and Triangles.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on Angles and Triangles.

Angles and Triangles – AIME I, 2012


Let triangle ABC be a right angled triangle with right angle at C. Let D and E be points on AB with D between A and E such that CD and CE trisect angle C. If \(\frac{DE}{BE}\)=\(\frac{8}{15}\), then tan B can be written as \(\frac{mp^\frac{1}{2}}{n}\) where m and n are relatively prime positive integers, and p is a positive integer not divisible by the square of any prime , find m+n+p.

  • is 107
  • is 18
  • is 840
  • cannot be determined from the given information

Key Concepts


Angles

Algebra

Triangles

Check the Answer


But try the problem first…

Answer: is 18.

Source
Suggested Reading

AIME I, 2012, Question 12

Geometry Vol I to Vol IV by Hall and Stevens

Try with Hints


First hint

Let CD=2a,then with angle bisector theorem of triangle we have for triangle CDB \(\frac{2a}{8}\)=\(\frac{CB}{15}\) then \(CB=\frac{15a}{4}\)

Second Hint

DF drawn perpendicular to BC gives CF=a, FD=\(a \times 3^\frac{1}{2}\), FB= \(\frac{11a}{4}\)

Final Step

then tan B = \(\frac{a \times 3^\frac{1}{2}}{\frac{11a}{4}}\)=\(\frac{4 \times 3^\frac{1}{2}}{11}\) then m+n+p=4+3+11=18.

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