Metrobank-MTAP-DepEd Math Challenge 2013 Individual Finals Third Year Category A Answer Key
•
•
•
Part I
69/ /20 1. 69
4. 432 cm2
7. 14
10. 10
2. 6x
5. 35, 46
27/ /7 8. 27
11.
9. 3π cm
12. 16
3.
− 8 cm (1, (1, −2)
6.
5 3
+
3 22
+
3 23
√
5 4
cm
√
13. 64 3 cm2
−2
14. 3/2 cm 15. 72◦
Part II 1. 2
(1 point) A = 23 + (1 point) B =
2 3
+
2 32
+
2 33
3
+
··· = + ··· =
− =3 2
1
1
2
2
− =1 (1 point) The sum of the series is A B = 2. 2. 3
1
1 3
−
2. 45◦
(1 point) In a regular hexagon, each interior angle is 120◦ so ∠ABN = ∠ABC
BC = 120◦ −90◦ = 30◦ . −∠N BC =
(1 point) ABN ABN is isosce isosceles les with with AB = N B (they are both equal to BC ). BC ). Ther Theref efor ore, e, 1 1 ◦ ◦ ◦ ◦ (180 ∠N AB) AB ) = 2 (180 30 ) = 75 . 2 (1 point) Finally, ∠F AN = ∠F AB ∠N AB = AB = 120◦ 75◦ = 45◦ .
−
3.
√
2+ 3 2π
or
1 π
+
−
−
∠N AB
−
√
3 2π
√
· · √ · √ · √
(1 point) N M Q is a 30◦ -60◦ -90◦ triangle. M Q = 1 and M and M N = 3. The area is 21 1 3= (1 point) N P Q is a 45◦ -45◦ -90◦ triangle. N P = 2 = QP . QP . The area is 21 2 2 = 1. √ 3 +1 1 3 (1 point) The required ratio is 2 2 = + . π (1) π 2π
√
√
√
3 2
4. 3
(1 point) S 12 12 =
12 2
[2a [2 a + 11d 11d] and S and S 5 = 25 [2a [2a + 4d 4d].
(1 point) The equation is S is S 12 12 = 6S 5 , i.e., 6(2a 6(2a + 11d 11d) = 15(2a 15(2a + 4d 4d) (1 point) This leads to
d = 3. 3. a
12 2
[2a [2 a + 11d 11d] = 6
=
⇒
·
5 2
[2a [2a + 4d 4d].
12 12a a + 66d 66d = 30 30a a + 60d 60d
=
⇒
6d = 18 18a a
.
=
5. 2 cm
(1 point) Letting r be the radius, and the points of tangency as indicated, we have BD = BE = r. Therefore, AD = 5 r and CE = 12 r.
−
−
(1 point) Since F is the point of tangency on AC , we have AF = AD = 5 Since the hypotenuse is 13, then (5 r) + (12 r) = 13.
−
(1 point) This equation yields r = 2.
−
− r and CF = C E = 12 − r.
Part III 1. 14/9
(1 point) The terms are a 1 = a, a4 = a + 3d and a 12 = a + 11d. (1 point) Since they form a geometric sequence,
a4 a1
=
a12 a4
, so (a + 3d)2 = a(a + 11d).
(2 points) Simplify: a2 + 6ad + 9d2 = a 2 + 11ad
(1 point) The required ratio is 2. 2
=
⇒
9d2 = 5ad
=
⇒
d 5 = . a 9
a2 a+d d 14 = = 1+ = . a1 a a 9
− √ 2
(2 points) A and C are both moved to G after the two foldings. If we let AE = C F = x, it follows that EG = F G = x and B E = BF = 2 x.
√ −
(2 points) EBF is an isosceles right triangle with leg 2( 2 x).
√ √ −
(1 point) Solving this yields x = 2 Alternative Solution
√ 2 − x and hypotenuse 2x.
Therefore, 2x =
− √ 2.
√ 2 and so BD = 2. (1 point) DAE coincides with DGE after the fold, so they are congruent. Thus, ∠ADE = ∠GDE . √ 2 − x AE BE x (2 point) DE is then an angle bisector in ADB. It follows that = , i.e., √ = . AD BD 2 2 √ (1 point) Solving this yields x = 2 − 2. (1 point) Note that AD =
3. 6
(3 points) It can be shown that ∠BAD = ∠CAD = θ. One way of showing this is by letting ∠CAP = θ. Since AP C CP D, it follows that ∠P CD = θ. Now, since P CD BAD (these are both right triangles with ∠ADB = ∠CDP ), it follows that ∠BAD = ∠P CD = θ.
∼
∼ √ √ √ (1 point) In the right triangle ABC , AB = x + 3 2 and therefore, AC = 2(x + 3 2). Alternatively, AC √ one may just note that 2AB without expressing this in terms of x. Since AD is an angle bisector of = x CD AC √ = = = 2. ABC , we have 3√ BD AB 2 √ √ (1 point) This yields x = 3 2 · 2 = 6.
