Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 12B Problems/Problem 13"

m (→‎Problem)
(→‎Solution)
 
(One intermediate revision by one other user not shown)
Line 8: Line 8:
 
Since the angles of Quadrilateral <math>ABCD</math> form an arithmetic sequence, we can assign each angle with the value <math>a</math>, <math>a+d</math>, <math>a+2d</math>, and <math>a+3d</math>. Also, since these angles form an arithmetic progression, we can reason out that <math>(a)+(a+3d)=(a+d)+(a+2d)=180</math>.
 
Since the angles of Quadrilateral <math>ABCD</math> form an arithmetic sequence, we can assign each angle with the value <math>a</math>, <math>a+d</math>, <math>a+2d</math>, and <math>a+3d</math>. Also, since these angles form an arithmetic progression, we can reason out that <math>(a)+(a+3d)=(a+d)+(a+2d)=180</math>.
  
−
For the sake of simplicity, lets rename the angles of each similar triangle. Lets call Angle <math>DBA</math> and Angle <math>DCB</math> Angle <math>1</math>. Also we rename Angle <math>ADB</math> and Angle <math>CBD</math> Angle <math>2</math>. Finally we rename Angles <math>BAD</math> and <math>BDC</math> Angle <math>3</math>.
+
For the sake of simplicity, lets rename the angles of each similar triangle. Let <math>\angle ADB = \angle CBD = \alpha</math>, <math>\angle DBA = \angle DCB = \beta</math>, <math>\angle CDB = \angle BAD = \gamma</math>.
  
−
Now we can rename the four angles of Quadrilateral <math>ABCD</math> as Angle <math>2</math>, Angle <math>1 + 2</math>, Angle <math>3</math>, and Angle <math>1 + 3</math>.
+
Now the four angles of <math>ABCD</math> are <math>\beta</math>, <math>\alpha + \beta</math>, <math>\gamma</math>, and <math>\alpha + \gamma</math>.
  
−
As for the similar triangles, whose Angles are equivalent, we can name them <math>y</math>, <math>y+b</math>, and <math>y+2b</math>. Therefore <math>y+y+b+y+2b=180</math> and <math>y+b=60</math>. 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.
+
As for the similar triangles, we can name their angles <math>y</math>, <math>y+b</math>, and <math>y+2b</math>. Therefore <math>y+y+b+y+2b=180</math> and <math>y+b=60</math>. Because these 3 angles are each equal to one of <math>\alpha, \beta, \gamma</math>, 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.
+
Now we we use trial and error. Let <math>\alpha = 60^{\circ}</math>. Then the angles of ABCD are <math>\beta</math>, <math>60^{\circ} + \beta</math>, <math>\gamma</math>, and <math>60^{\circ} + \gamma</math>. Since these four angles add up to 360, then <math>\beta + \gamma= 120</math>. If we list them in increasing value, we get <math>\beta</math>, <math>\gamma</math>, <math>60^{\circ} + \beta</math>, <math>60^{\circ} + \gamma</math>. Note that this is the only sequence that works because the common difference is less than 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, <math>\alpha, \beta, \gamma</math> 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, <math>\boxed{\textbf{(D) }240}</math> is the correct answer.
+
If we apply the same reasoning to <math>\beta</math> and <math>\gamma</math>, 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 ==

Latest revision as of 11:32, 8 January 2021

Problem

The internal angles of quadrilateral $ABCD$ form an arithmetic progression. Triangles $ABD$ and $DCB$ are similar with $\angle DBA = \angle DCB$ and $\angle ADB = \angle CBD$. Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the two largest angles of $ABCD$?

$\textbf{(A)}\ 210 \qquad \textbf{(B)}\ 220 \qquad \textbf{(C)}\ 230 \qquad \textbf{(D)}\ 240 \qquad \textbf{(E)}\ 250$

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. Let $\angle ADB = \angle CBD = \alpha$, $\angle DBA = \angle DCB = \beta$, $\angle CDB = \angle BAD = \gamma$.

Now the four angles of $ABCD$ are $\beta$, $\alpha + \beta$, $\gamma$, and $\alpha + \gamma$.

As for the similar triangles, we can name their angles $y$, $y+b$, and $y+2b$. Therefore $y+y+b+y+2b=180$ and $y+b=60$. Because these 3 angles are each equal to one of $\alpha, \beta, \gamma$, we know that one of these three angles is equal to 60 degrees.

Now we we use trial and error. Let $\alpha = 60^{\circ}$. Then the angles of ABCD are $\beta$, $60^{\circ} + \beta$, $\gamma$, and $60^{\circ} + \gamma$. Since these four angles add up to 360, then $\beta + \gamma= 120$. If we list them in increasing value, we get $\beta$, $\gamma$, $60^{\circ} + \beta$, $60^{\circ} + \gamma$. Note that this is the only sequence that works because the common difference is less than 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, $\alpha, \beta, \gamma$ 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 $\beta$ and $\gamma$, we would get the sum of the highest two angles as 220, which works but is lower than 240. Therefore, $\boxed{\textbf{(D) }240}$ 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_13&oldid=141723"