Difference between revisions of "2003 AIME I Problems/Problem 10"
(→Solution 6 (Very simple)) |
Trigonatops (talk | contribs) (Add protractor solution (which I used back in 2003)) |
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<cmath>\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta</cmath> | <cmath>\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta</cmath> | ||
− | and multiplying through by 2 and applying the [ | + | and multiplying through by 2 and applying the [https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities#Double-angle_identities| double angle formulas] gives |
<cmath>\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta</cmath> | <cmath>\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta</cmath> | ||
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By inspection, <math>x=14^\circ</math> works, so the answer is <math>180-83-14= \boxed{083}</math> | By inspection, <math>x=14^\circ</math> works, so the answer is <math>180-83-14= \boxed{083}</math> | ||
+ | === Solution 6 === | ||
− | === | + | Let <math>\angle{APC} = \theta^{\circ}</math> |
+ | Using sine rule on <math>\triangle{APB}, \triangle{APC}</math>, letting <math>AP=d</math> we get : | ||
+ | <math>\frac{d}{1} = \frac{\sin{7^{\circ}}}{\sin{150^{\circ}}} = 2\sin{7^{\circ}}= \frac{\sin{14^{\circ}}}{\cos{7^{\circ}}}= \frac{\sin{14^{\circ}}}{\sin{83^{\circ}}}= \frac{\sin{(97-\theta)^{\circ}}}{\sin{\theta^{\circ}}}</math> | ||
+ | Simplifying, we get that | ||
+ | <math>\cos{(14-\theta)^{\circ}}-\cos{(14+\theta)^{\circ}}=\cos{(14-\theta)^{\circ}}-\cos{(180-\theta)^{\circ}},</math> from where <math>\cos{(14-\theta)^{\circ}}=\cos{(180-\theta)^{\circ}}</math> | ||
+ | Simplifying more, we get that <math>\sin{97^{\circ}} \cdot \sin{(\theta-83)^{\circ}} = 0</math>, so <math>\theta = 83^{\circ}</math> | ||
+ | NOTE: The simplifications were carried out by the product-to-sum and sum-to-product identities | ||
+ | ~Prabh1512 | ||
− | + | === Solution 7 === | |
+ | |||
+ | Because protractors are allowed on the AIME, it is practical to solve this problem by carefully drawing the picture and measuring the angle. | ||
== See also == | == See also == |
Latest revision as of 02:08, 30 September 2024
Problem
Triangle is isosceles with and Point is in the interior of the triangle so that and Find the number of degrees in
Contents
Solutions
Solution 1
Take point inside such that and .
. Also, since and are congruent (by ASA), . Hence is an equilateral triangle, so .
Then . We now see that and are congruent. Therefore, , so .
Solution 2
From the givens, we have the following angle measures: , . If we define then we also have . Then apply the Law of Sines to triangles and to get
Clearing denominators, evaluating and applying one of our trigonometric identities to the result gives
and multiplying through by 2 and applying the double angle formulas gives
and so ; since , we must have , so the answer is .
Solution 3
Without loss of generality, let . Then, using the Law of Sines in triangle , we get , and using the sine addition formula to evaluate , we get .
Then, using the Law of Cosines in triangle , we get , since . So triangle is isosceles, and .
Solution 4
Note: A diagram would be much appreciated; I cannot make one since I'm bad at asymptote. Also, please make this less cluttered :) ~tauros
First, take point outside of so that is equilateral. Then, connect , , and to . Also, let intersect at . , , and (trivially) , so by SAS congruence. Also, , so , and , making also equilateral. (it is isosceles with a angle) by SAS (, , and ), and by SAS (, , and ). Thus, is isosceles, with . Also, , so .
Solution 5 (Ceva)
Noticing that we have three concurrent cevians, we apply Ceva's theorem:
using the fact that and we have:
By inspection, works, so the answer is
Solution 6
Let Using sine rule on , letting we get : Simplifying, we get that from where Simplifying more, we get that , so NOTE: The simplifications were carried out by the product-to-sum and sum-to-product identities ~Prabh1512
Solution 7
Because protractors are allowed on the AIME, it is practical to solve this problem by carefully drawing the picture and measuring the angle.
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.