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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

Angles

Algebra

Triangles

## Check the Answer

But try the problem first…

Source

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.