Soal Fim.docx

1. Find arctan 2+arctan 3 in degrees Solution: Let

x=arctan 2+arctan 3 . Then the tangent-sum formula gives:

2 3 arctan ¿ ¿ ¿ ¿ 2 3 arctan ¿ ¿ ¿ arctan ¿ tan ⁡¿ arctan ¿+ tan ⁡¿ ¿ tan ¿ tan x=¿ There are two places where tan x=−1:135 ° ,∧325° . 45° =arctan 1<arctan 2< arctan 3<90 ° . °

So, 90 < arctan 2+arctan 3<180 arctan 2+arctan 3=135

°

3 arctan 2+arctan ¿ , then . Since ¿ tan ⁡¿

°

2. in the following diagram, which is not drawn to scale, the shaded area is 32 π . The circle has radius BC =BA =6 . Find the measure of ∠ ABC .

Solution: The area of the circle is 36 π .

The area of the triangle is 4 π =( 1/2 ) 6.6 sin (∠ ABC ) So sin(∠ ABC )=2 π /9 . Thus the angle is arcsin ⁡(

2π ) 9

3. An octagon in the plane is symmetric about the x -axis, the y -axis, and the line whose equations is y=x . If (1, √ 3) is a vertex of the octagon, find its area Solution: The vertices of the octagon are ( ± 1,± √ 3 ) and ( ± √3 , ± 1 ) . The area can be computed by computing the area of the enclosing square that is 2 √ 3 ×2 √ 3 and subtracting the area of the four right isosceles triangles or, equivalently, the area of two squares that are 2 2 √ 3−1 × √ 3−1 . The area is ( 2 √3 ) −2 ( √3−1 ) =12−2 ( 4−2 √3 ) =4 +4 √ 3

△ ABC , AC=13, BC =15 and the area of △ ABC=84 . If CD=7,CE =13 and p the area of △ CDE can be represented as where p and q are relatively prime q positive integer, find q .

4. In

Solution: The area of

△ ABC

is

1 1 . AC . BC . sin C= .13. 15. sin C=84 2 2

The area of

△ CDE

is

1 1 .CD .CE . sin C= .7 . 13 . sinC . Let [ ABC ] be the area of 2 2

△ ABC

and [ CDE ]

[ CDE ] =

be the area of

△ CDE . Then

[ CDE ] 7.13 7 = = . So [ ABC ] 13.15 15

7 7 196 [ ABC ] = . 84= . So, q=5 15 15 5

5. A circle centered at O is inscribed in the equilateral triangle ABC . Line segment DE is tangent to the circle, is perpendicular to AB , and intersects AB and AC at points D and E respectively. (See the sketch below) If AD =1 inch, what is the length of one of the triangle’s sides? Sumber: BC exam Texas A&M High School Math Contest. October 24, 2015

Solution: let G be the point on the circle at which DE is tangent. Draw a straight line from O to F (the point where the circle is tangent to AB ¿ and from O to G . The quadrilateral GOFD is a square, with side length equal to the radius of the circle. Since triangle ABC is equilateral, the radius of the circle, r and the triangle’s side length, s satisfy the equation r=

s 2 √3

Moreover the point

F

s=2. AF=2 ( r +1 ) =2

bisects s

( 2√ 3 +1)

Solving this equation for s s=

2 √3 =3+ √3 √ 3−1

AB . Thus

inches

we have

6. α ∧β

are two angles from the interval ¿ such that the following two relations hold

{

3 sin2 α +2 sin2 β=1, 3 sin ( 2 α )−2 sin ( 2 β )=0

Find α + 2 β Solution: From the first equation 3 sin2 α =1−2 sin 2 β =cos ⁡( 2 β ) . From the second equation 3 sin ( 2 β ) = sin ⁡( 2α ) . 2 Therefore, 3 2 2 2 cos ( α +2 β ) =cos α cos(2 β)−sin α sin ( 2 β )=cos α .3 sin α −sin α . sin ( 2 α )=3 cos α sin α−3 cos α sin α =0 2 Since α ∧β

[ ) 0,

3π 2

belong the interval

[ ) 0,

π 2

, we have that α + 2 β

belong to the interval

. therefore the fact that cos ( α +2 β ) =0 implies that ( α + 2 β )=

π 2