Difference between revisions of "2017 AMC 12A Problems/Problem 24"
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By multiplying the <math>2</math> equations we get <math>XF \cdot XG = \frac{XE}{XA} \cdot XD \cdot XB</math> | By multiplying the <math>2</math> equations we get <math>XF \cdot XG = \frac{XE}{XA} \cdot XD \cdot XB</math> | ||
− | <math>\because</math> <math> | + | <math>\because</math> <math>YE \parallel AD</math>, <math>\quad \therefore</math> <math>\triangle EYX \sim \triangle ADX</math>, <math>\quad \frac{XD}{XY} = \frac{XA}{XE}, \quad XD \cdot XE = XA \cdot XY, \quad XD = \frac{XA \cdot XY}{XE}</math> |
− | By substitution, <math>XF \cdot XG = XY \cdot XB = \frac{4}{9} BD \cdot \frac{3}{4} BD = \frac{BD^2}{3}</math> | + | By substitution, <math>XF \cdot XG = \frac{XE}{XA} \cdot \frac{XA \cdot XY}{XE} \cdot XB = XY \cdot XB = \frac{4}{9} BD \cdot \frac{3}{4} BD = \frac{BD^2}{3}</math> |
Let <math>a = AB</math>, <math>b = BC</math>, <math>c = CD</math>, <math>d = AD</math>, <math>p = AC</math>, and <math>q = BD</math> | Let <math>a = AB</math>, <math>b = BC</math>, <math>c = CD</math>, <math>d = AD</math>, <math>p = AC</math>, and <math>q = BD</math> | ||
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<cmath>[ABC] = \frac12 \cdot ab \cdot \sin B, \quad [ACD] = \frac12 \cdot cd \cdot \sin D = \frac12 \cdot cd \cdot \sin B</cmath> | <cmath>[ABC] = \frac12 \cdot ab \cdot \sin B, \quad [ACD] = \frac12 \cdot cd \cdot \sin D = \frac12 \cdot cd \cdot \sin B</cmath> | ||
+ | |||
+ | <cmath>[ABCD] = [ABD] + [BCD] = \frac12 \cdot ad \cdot \sin A + \frac12 \cdot bc \cdot \sin A = \frac12 (ad + bc) \sin A</cmath> | ||
<cmath>[ABCD] = [ABC] + [ACD] = \frac12 \cdot ab \cdot \sin B + \frac12 \cdot cd \cdot \sin B = \frac12 (ab + cd) \sin B</cmath> | <cmath>[ABCD] = [ABC] + [ACD] = \frac12 \cdot ab \cdot \sin B + \frac12 \cdot cd \cdot \sin B = \frac12 (ab + cd) \sin B</cmath> |
Latest revision as of 02:24, 7 October 2023
Contents
Problem
Quadrilateral is inscribed in circle and has side lengths , and . Let and be points on such that and . Let be the intersection of line and the line through parallel to . Let be the intersection of line and the line through parallel to . Let be the point on circle other than that lies on line . What is ?
Diagram
~raxu, put in by fuzimiao2013
Solution 1
Using the given ratios, note that
By AA Similarity, with a ratio of and with a ratio of , so .
Now we find the length of . Because the quadrilateral is cyclic, we can simply use the Law of Cosines. By Power of a Point, . Thus
-solution by FRaelya
Solution 2
We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point. Let be the intersection of and . First, from being a cyclic quadrilateral, we have that , . Therefore, , , and , so we have , , and . By Ptolemy's Theorem, Thus, . Then, by Power of a Point, . So, . Next, observe that , so . Also, , so . We can compute after noticing that and that . So, . Then, .
Multiplying our equations for and yields that
Solution 3
Denote to be the intersection between line and circle . Note that , making . Thus, is a cyclic quadrilateral. Using Power of a Point on gives .
Since and , . Using Power of a Point on again, . Plugging in gives: By Law of Cosines, we can find , as in Solution 1. Now, and , making . This gives us as a result.
-Solution by sml1809
Note
You could have also got the relation as follows: From the similarities, . PoP on gives . Plugging in and gives implying that .
~sml1809
Solution 4
, ,
By Power of a Point,
By multiplying the equations we get
, ,
By substitution,
Let , , , , , and
By Ptolemy's theorem,
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=JdERP0d0W64&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=4 - AMBRIGGS
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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.