HighFour Mathematics Category D: Grades 11 – 12 Round 6 Thursday, February 16, 2017 The use of calculator is required. Answer #1 Explanation: Answer #2 Explanation: 20.78 The largest possible area comes with the equilateral triangle. After drawing the equilateral triangle in the circle and using SOH-­‐CAH-­‐TOA to find the length of its sides, we find that the area of such inscribed triangle is 12 3 ≈ 20.78. 1.414 Taking the base 𝑥 logarithm of both sides of the equation gives log - 𝑥
-.
..
.
= log - 4 ⟹ 𝑥
-.
..
.
= log - 4 = 4 Solving this equation for 𝑥 gives 𝑥 2 = 4 ⟹ 𝑥 = 2 ≈ 1.414. Answer #3 Explanation: 172.8 degrees 4
The number of diagonals of a polygon with 𝑛 sides is 𝑛 𝑛 − 3 . As 5
4
such, we have 1175 = 𝑛 𝑛 − 3 . Solving this for 𝑛 gives the equation 5
5
𝑛 − 3𝑛 − 2350 = 0 ⟹ 𝑛 − 50 𝑛 + 47 = 0 ⟹ 𝑛 = 50 given that 𝑛 > 0. Then, the measure of interior angles of a regular polygon with 𝑛 sides is Answer #4 Explanation: 4:; <=5
<
. Substituting 𝑛 = 50 gives 172.8 degrees. 72 km/h Since either time he arrives at is 5 minutes from the desired time, the average speed he should have is the harmonic mean of 60 and 90: 2
2 60×90
=
= 72 4
4
60 + 90
+
>;
?;
Therefore, Jake should ride his scooter at an average speed of 72 km/h to arrive exactly on time. HighFour Mathematics Category D: Grades 11 – 12 Round 6 Thursday, February 16, 2017 The use of calculator is required. Answer #5 Explanation: 0.25 Let 𝑎 and 𝑟 represent the first term and the common ratio of the geometric sequence, respectively. We are given that 𝑎 + 𝑎𝑟 = 2.5 and :
F
=
. Expressing 𝑎 in the first equation and then substituting it into the E
4=G
second equation gives the following 5.H
Answer #6 Explanation: 8
1
= 4IG ⟹ 𝑟 = = 0.25 3 1−𝑟
4
25% 4
The area of the triangle is 𝑏ℎ. After the change, considering that the area 5
4
4
5
5
must remain the same, we have 𝑏ℎ =
0.8𝑏 𝑥ℎ , where 𝑥 represents 4
the percent of the new height. Solving this for 𝑥 gives 𝑥 =
= 1.25. ;.:
Therefore, the height must be increased by 25% to keep the same area. Answer #7 Explanation: Answer #8 Explanation: 8 meters By the Pythagorean triple 7-­‐24-­‐25, the top of the ladder is 24 meters high on the wall. When it slides down 4 meters, it will be 20 meters high on the wall. Using the other Pythagorean triple 15-­‐20-­‐25, where 20 is the height the ladder reaches on the wall, and 25 is the length of the ladder, we have that the foot of the ladder is 15 meters away from the wall. Therefore, the foot of the ladder slides 15 𝑚 − 7 𝑚 = 8 𝑚 further away from the wall. 428.45 Let 𝑑 , 𝑡4 , and 𝑡5 represent the depth of the cliff, the number of seconds the rock took to reach the bottom, and the number of seconds the sound took to travel back up, respectively. We know that 𝑡4 + 𝑡5 = 10.6 and we can write the following equations: 𝑑 = 4.9𝑡45 ⟹ 𝑡4 =
P
2.?
and 𝑑 = 343𝑡5 ⟹ 𝑡5 =
P
E2E
Thus, 𝑑
𝑑
+
= 10.6 4.9 343
Solving this equation for 𝑑 gives 𝑑 ≈ 428.45 meters, rounded to two decimal places. HighFour Mathematics Category D: Grades 11 – 12 Round 6 Thursday, February 16, 2017 The use of calculator is required. Answer #9 12 Explanation: The “handshake equation” to use is ℎ =
, where ℎ stands for the 5
number of handshakes and 𝑛 stands for the number of people exchanging < <=4
< <=4
handshakes. Thus, 66 =
⟹ 122 = 𝑛 𝑛 − 1 ⟹ 𝑛 = 12 . 5
Therefore, there were 12 people exchanging handshakes at the party. Answer #10 Explanation: Answer #11 Explanation: 16 Leaving two empty seats between any two people means that any person taking a seat must be seated next to someone being seated already. Therefore, the smallest number of people that can be seated this way is one-­‐third of all seats around the table, which is 16. 3 days Let 𝑎, 𝑏, and 𝑐 represent the percent of chimneys Aldo, Brent, and Calvin builds in a day, respectively. Writing what is known as equations we have 4
𝑎 + 𝑏 = chimneys per day 5
4
𝑏 + 𝑐 = chimneys per day 2
H
𝑐 + 𝑎 = chimneys per day 45
Adding these equations together and then simplifying gives R
𝑎 + 𝑏 + 𝑐 = chimneys per day 45
Answer #12 Explanation: R
4
4
Therefore, as 𝑎 = 𝑎 + 𝑏 + 𝑐 − 𝑏 + 𝑐 = − = , it takes Aldo 3 45
2
E
days to build a chimney on his own. 30.545 minutes The minute hand moves at a rate of 360° per hour, while the hour hand moves at 30° per hour. Therefore, the minute hand catches up to the hour hand at rate of 360° − 30° = 330° per hour. Therefore, it will take 54;°=:2°
45>°
=
hour for the two hands of the clock to make an 84° angle. EE;°
EE;°
54;°I:2°
5?2°
It will also take =
hour after 7:00 for the hands to make an EE;°
EE;°
84° angle for the second time. Converting these to minutes we see that it 4;
H
will take 22.909 = 22 minutes and 53. 45 = 53 minutes for the two 44
44
hands to make the 84° angle. The difference between these two times then is 30.545 minutes, rounded to three decimal places. HighFour Mathematics Category D: Grades 11 – 12 Round 6 Thursday, February 16, 2017 The use of calculator is required. Answer #13 Explanation: Answer#14 Explanation: Answer #15 Explanation: 4.25 minutes At 120 steps per minute and 70 cm per step, Jack walks at a rate of 120×70 = 8400 cm per minute. In 15 minutes, he walks 126000 cm, which is the length of the route to school. Jane at 100 steps per minute and 50 cm per step, walks at a rate of 100×50 = 5000 cm per minute. 45>;;;
Therefore, it takes her = 25.2 minutes to get to school, which is H;;;
10. 2 minutes longer than it takes Jack. 11.18 cm Let the edges be represented by 𝑎, 𝑏, and 𝑐 . The sum of the areas of the faces is 2× 𝑎𝑏 + 𝑎𝑐 + 𝑏𝑐 = 164 , which gives 𝑎𝑏 + 𝑎𝑐 + 𝑏𝑐 = 82 . Similarly, the sum of all edges is 4× 𝑎 + 𝑏 + 𝑐 = 68, which in turn gives 𝑎 + 𝑏 + 𝑐 = 17. Given that 𝑎 + 𝑏 + 𝑐 5 = 𝑎5 + 𝑏 5 + 𝑐 5 + 2𝑎𝑏 + 2𝑎𝑐 + 2𝑏𝑐 , we find that 𝑎5 + 𝑏 5 + 𝑐 5 = 175 − 164 = 125 .Therefore the length of the interior diagonal is 𝑎5 + 𝑏 5 + 𝑐 5 = 125 ≈ 11.18 , rounded to two decimal places. 15% The volume of the rectangular prism is 𝑉 = 8 𝑐𝑚×12 𝑐𝑚×15 𝑐𝑚 =
1440 𝑐 𝑚E . The combined volume of the eight cubes removed is 8× 3 𝑐𝑚 3 𝑐𝑚 3 𝑐𝑚 = 216 𝑐 𝑚E . Therefore, the removed volume was Answer #16 Explanation: 54> U V W
422; U V W
×100% = 15% of the original rectangular prism. 216 Let 𝑛 represent the smallest of the 16 numbers. Thus, the sum of these 4H×4>
16 numbers is 𝑛 + 𝑛 + 1 + ⋯ + 𝑛 + 15 = 16𝑛 +
, which can be 5
factorized as 8 2𝑛 + 15 . As this expression is a perfect cube, and 8 is a perfect cube, 2𝑛 + 15 must also be a perfect cube. The smallest possible value for this expression is 27, from the list of perfect cubes 1, 8, 27, 64, and so forth. Therefore, 2𝑛 + 15 = 27 ⟹ 𝑛 = 6 , giving the smallest possible perfect cube 6E = 216. HighFour Mathematics Category D: Grades 11 – 12 Round 6 Thursday, February 16, 2017 The use of calculator is required. Answer #17 Explanation: 36 Notice that there are the five triangles are overlapping in four small 4
triangles, each of which with an area of of the larger triangles. As such, 2
the area of the fixture is the same as the sum of the areas of four non-­‐
overlapping triangles. Z
Given that the the base of the triangle is 2 27, its height is Z
2 27
5
1 Z
−
2 27
2
5
=
4 27 −
1
4 27 =
4
3 27 =
9 3
Z
= 3 3 Therefore, the area of four non-­‐overlapping triangles with sides 2 27 is 4×
Answer #18 Explanation: Z
4
5
Z
Z
2 27 3 3
Z
= 4× 3 81 = 4×3×3 = 36. 96.904 meters Let us convert the speed of sound into m/s: 1234800
1234.8 𝑘𝑚/ℎ =
𝑚/𝑠 = 343𝑚/𝑠 60
Let 𝑑 represent the distance between the house and the mouse at the time when it heard the blast and 𝑡 represent the time it took the blast to be heard by the mouse. Thus, the following equation can be written: 𝑑 = 3.2× 30 + 𝑡 = 343×𝑡 Solving this for 𝑡 , gives 𝑡 = 0.2825191 …, which means the mouse has been running for 30.2825191… seconds when it heard the blast. In this time, it ran 𝑑 = 3.2×30.2825191 … ≈ 96.904 meters, rounded to three decimal places. HighFour Mathematics Category D: Grades 11 – 12 Round 6 Thursday, February 16, 2017 The use of calculator is required. Answer #19 Explanation: 13 Let 𝑛 represent the number of sides of Chris’ polygon and 𝑂4 , … , 𝑂H and 𝐴4 , 𝐴5 , … , 𝐴<=H represent its obtuse and acute interior angles, respectively. For the five obtuse angles we have 90 < 𝑂4 , … , 𝑂H < 270 and for the rest of the acute angles we have 0 < 𝐴4 , 𝐴5 , … , 𝐴<=H < 90. Thus, 450 < 𝑂4 + ⋯ + 𝑂H + 𝐴4 + ⋯ + 𝐴<=H < 5×270 + 90× 𝑛 − 5 . Given that the sum of all interior angles is 180°× 𝑛 − 2 , we have the following equation: Answer #20 Explanation: 450 < 180 𝑛 − 2 < 1350 + 90 𝑛 − 5 Simplifying this inequality gives 810 < 180𝑛 < 1260 + 90𝑛 ⟹ 4.5 < 𝑛 < 14 Therefore, Chris’ polygon can have 5, 6, …, 13 sides. 15.7 Let 𝑛 represent the number of students in the smaller class, which means there are 1.5𝑛 students in the larger class. Thus, the following equations can be established, where 𝑆4 and 𝑆5 is the sum of the scores in the smaller class and in the larger class, respectively: 𝑆4
= 17.5,
𝑛
𝑆5
= 14.5,
1.5𝑛
𝑆4 + 𝑆5
= 𝐴 𝑛 + 1.5𝑛
Expressing 𝑛 from the first two equations and then substituting them into the third gives 17.5𝑛 + 21.75𝑛
= 𝐴 ⟹ 𝐴 = 15.7 2.5𝑛
Therefore, the average score of the two classes combined was 15.7 points.