Difference between revisions of "2020 AIME I Problems/Problem 1"
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If we set <math>\angle{BAC}</math> to <math>x</math>, we can find all other angles through these two properties: | If we set <math>\angle{BAC}</math> to <math>x</math>, we can find all other angles through these two properties: | ||
1. Angles in a triangle sum to <math>180^{\circ}</math>. | 1. Angles in a triangle sum to <math>180^{\circ}</math>. | ||
− | 2. The base angles of an | + | 2. The base angles of an isosceles triangle are congruent. |
Now we angle chase. <math>\angle{ADE}=\angle{EAD}=x</math>, <math>\angle{AED} = 180-2x</math>, <math>\angle{BED}=\angle{EBD}=2x</math>, <math>\angle{EDB} = 180-4x</math>, <math>\angle{BDC} = \angle{BCD} = 3x</math>, <math>\angle{CBD} = 180-6x</math>. Since <math>AB = AC</math> as given by the problem, <math>\angle{ABC} = \angle{ACB}</math>, so <math>180-4x=3x</math>. Therefore, <math>x = 180/7^{\circ}</math>, and our desired angle is <cmath>180-4\left(\frac{180}{7}\right) = \frac{540}{7}</cmath> for an answer of <math>\boxed{547}</math>. | Now we angle chase. <math>\angle{ADE}=\angle{EAD}=x</math>, <math>\angle{AED} = 180-2x</math>, <math>\angle{BED}=\angle{EBD}=2x</math>, <math>\angle{EDB} = 180-4x</math>, <math>\angle{BDC} = \angle{BCD} = 3x</math>, <math>\angle{CBD} = 180-6x</math>. Since <math>AB = AC</math> as given by the problem, <math>\angle{ABC} = \angle{ACB}</math>, so <math>180-4x=3x</math>. Therefore, <math>x = 180/7^{\circ}</math>, and our desired angle is <cmath>180-4\left(\frac{180}{7}\right) = \frac{540}{7}</cmath> for an answer of <math>\boxed{547}</math>. |
Revision as of 16:56, 28 May 2020
Contents
[hide]Problem
In with point lies strictly between and on side and point lies strictly between and on side such that The degree measure of is where and are relatively prime positive integers. Find
Solution 1
If we set to , we can find all other angles through these two properties: 1. Angles in a triangle sum to . 2. The base angles of an isosceles triangle are congruent.
Now we angle chase. , , , , , . Since as given by the problem, , so . Therefore, , and our desired angle is for an answer of .
See here for a video solution: https://youtu.be/4e8Hk04Ax_E
Solution 2
Let be in degrees. . By Exterior Angle Theorem on triangle , . By Exterior Angle Theorem on triangle , . This tells us and . Thus and we want to get an answer of .
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25
(Almost Mirrored)
See here for a video solution:
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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