Difference between revisions of "2013 AMC 12B Problems/Problem 13"
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Now we we use trial and error to find out which of these 3 angles has a value of 60. If we substitute 60 degrees into Angle 1. This would cause the angle values of ABCD to be Angle 2, 60+Angle 2, Angle 3, and 60 + Angle 3. Since these four angles add up to 360, then Angle 2 + Angle 3 = 120. If we list them in increasing value, we get Angle 2, Angle 3, 60 + Angle 2, 60+Angle 3. Note that this is the only sequence that works because the common difference between each term cannot equal or exceed 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, Angle 1, 2, and 3, the angles of both similar triangles, also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 which is an answer choice. | Now we we use trial and error to find out which of these 3 angles has a value of 60. If we substitute 60 degrees into Angle 1. This would cause the angle values of ABCD to be Angle 2, 60+Angle 2, Angle 3, and 60 + Angle 3. Since these four angles add up to 360, then Angle 2 + Angle 3 = 120. If we list them in increasing value, we get Angle 2, Angle 3, 60 + Angle 2, 60+Angle 3. Note that this is the only sequence that works because the common difference between each term cannot equal or exceed 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, Angle 1, 2, and 3, the angles of both similar triangles, also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 which is an answer choice. | ||
| − | If we apply the same reasoning to Angles 2 and 3, we would get the sum of the highest two angles as 220, which works but is lower than 240. Therefore, 240 | + | If we apply the same reasoning to Angles 2 and 3, we would get the sum of the highest two angles as 220, which works but is lower than 240. Therefore, <math>\boxed{\textbf{(D) }240}</math> is the correct answer. |
== See also == | == See also == | ||
Revision as of 20:30, 7 November 2013
Problem
The internal angles of quadrilateral 
form an arithmetic progression. Triangles 
and 
are similar with 
and 
. Moreover, the angles in each of these two triangles also form an arithemetic progression. In degrees, what is the largest possible sum of the two largest angles of ?
Solution
Since the angles of Quadrilateral ABCD form an arithmetic sequence, we can assign each angle with the value a, a+d, a+2d, and a+3d. Also, since these angles form an arithmetic progression, we can reason out that (a)+(a+3d)=(a+d)+(a+2d)=180.
For the sake of simplicity, lets rename the angles of each similar triangle. Lets call Angle DBA and Angle DCB Angle 1. Also we rename Angle ADB and Angle CBD Angle 2. Finally we rename Angles BAD and BDC Angle 3.
Now we can rename the four angles of Quadrilateral ABCD as Angle 2, Angle 1 + 2, Angle 3, and Angle 1 + 3.
As for the similar triangles, whose Angles are equivalent, we can name them y, y+b, and y+2b. Therefore y+y+b+y+2b=180 and y+b=60. Because these 3 angles are each equal to one of the angles we named Angles 1, 2, and 3, we know that one of these three angles is equal to 60 degrees.
Now we we use trial and error to find out which of these 3 angles has a value of 60. If we substitute 60 degrees into Angle 1. This would cause the angle values of ABCD to be Angle 2, 60+Angle 2, Angle 3, and 60 + Angle 3. Since these four angles add up to 360, then Angle 2 + Angle 3 = 120. If we list them in increasing value, we get Angle 2, Angle 3, 60 + Angle 2, 60+Angle 3. Note that this is the only sequence that works because the common difference between each term cannot equal or exceed 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, Angle 1, 2, and 3, the angles of both similar triangles, also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 which is an answer choice.
If we apply the same reasoning to Angles 2 and 3, we would get the sum of the highest two angles as 220, which works but is lower than 240. Therefore, 
is the correct answer.
See also
| 2013 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
