Difference between revisions of "2021 Fall AMC 12B Problems/Problem 8"
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Set the vertex angle to be <math>a</math>, we derive the equation: | Set the vertex angle to be <math>a</math>, we derive the equation: | ||
| − | <math>x^2=4(\frac{1}{2}x^2\sin(a))</math> | + | <math>x^2=4\left(\frac{1}{2}x^2\sin(a)\right)</math> |
<math>\sin(a)=\frac{1}{2}</math> | <math>\sin(a)=\frac{1}{2}</math> | ||
| Line 20: | Line 20: | ||
~Wilhelm Z | ~Wilhelm Z | ||
| − | {{AMC12 box|year=2021 Fall|ab= | + | == Solution 2 == |
| + | Denote by <math>a</math> the length of each congruent side. Denote by <math>\theta</math> the degree measure of each acute angle. | ||
| + | Denote by <math>\phi</math> the degree measure of the obtuse angle. | ||
| + | |||
| + | Hence, this problem tells us the following relationship: | ||
| + | <cmath> | ||
| + | \[ | ||
| + | a^2 = 2 a \cos \theta \cdot 2 a \sin \theta . | ||
| + | \] | ||
| + | </cmath> | ||
| + | |||
| + | Hence, | ||
| + | <cmath> | ||
| + | \begin{align*} | ||
| + | 1 & = 2 \cdot 2 \sin \theta \cos \theta \\ | ||
| + | & = 2 \sin 2 \theta \\ | ||
| + | & = 2 \sin \left( 180^\circ - 2 \theta \right) \\ | ||
| + | & = 2 \sin \phi . | ||
| + | \end{align*} | ||
| + | </cmath> | ||
| + | |||
| + | Hence, <math>\phi = 150^\circ</math>. | ||
| + | |||
| + | Therefore, the answer is <math>\boxed{\textbf{(D) }150}</math>. | ||
| + | |||
| + | ~Steven Chen (www.professorchenedu.com) | ||
| + | |||
| + | |||
| + | ==Video Solution (Under 2 min!)== | ||
| + | https://youtu.be/nXnS6pn8iJc | ||
| + | |||
| + | <i>~Education, the Study of Everything</i> | ||
| + | |||
| + | ==Video Solution by TheBeautyofMath== | ||
| + | https://www.youtube.com/watch?v=4qgYrCYG-qw&t=795 | ||
| + | |||
| + | ~IceMatrix | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2021 Fall|ab=B|num-a=9|num-b=7}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:07, 10 July 2023
Contents
Problem
The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?
Solution 1 (Area)
Let the lengths of the two congruent sides of the triangle be 
, then the product desired is 
.
Notice that the product of the base and twice the height is 
times the area of the triangle.
Set the vertex angle to be 
, we derive the equation:
As the triangle is obtuse, 
only. We get 
~Wilhelm Z
Solution 2
Denote by the length of each congruent side. Denote by 
the degree measure of each acute angle.
Denote by 
the degree measure of the obtuse angle.
Hence, this problem tells us the following relationship:
![\[ a^2 = 2 a \cos \theta \cdot 2 a \sin \theta . \]](http://latex.artofproblemsolving.com/c/3/2/c328ec8657d0713415c84b2d34b0e134f138e24e.png)
Hence,
Hence, 
.
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
Video Solution (Under 2 min!)
~Education, the Study of Everything
Video Solution by TheBeautyofMath
https://www.youtube.com/watch?v=4qgYrCYG-qw&t=795
~IceMatrix
See Also
| 2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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. 
