Difference between revisions of "2016 AIME II Problems/Problem 10"
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==Solution 2 (Projective Geometry)== | ==Solution 2 (Projective Geometry)== | ||
− | Projecting through <math>C</math> we have <cmath>\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}</cmath> which easily gives <math>ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{ | + | [[File:2016 AIME II 10c.png|400px|right]] |
+ | Projecting through <math>C</math> we have <cmath>\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}</cmath> which easily gives <math>ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{043}</math>. | ||
==Solution 3== | ==Solution 3== | ||
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==Solution 4 == | ==Solution 4 == | ||
− | Extend <math>\overline{AB}</math> past <math>B</math> to point <math>X</math> so that <math>CPTX</math> is cyclic. Then, by Power of a Point on <math>CPTX</math>, <math>(CQ)( | + | Extend <math>\overline{AB}</math> past <math>B</math> to point <math>X</math> so that <math>CPTX</math> is cyclic. Then, by Power of a Point on <math>CPTX</math>, <math>(CQ)(QT) = (PQ)(QX)</math>. By Power of a Point on <math>CATB</math>, <math>(CQ)(QT) = (AQ)(QB) = 42</math>. Thus, <math>(PQ)(QX) = 42</math>, so <math>BX = 8</math>. |
By the Inscribed Angle Theorem on <math>CPTX</math>, <math>\angle SCT = \angle BXT</math>. By the Inscribed Angle Theorem on <math>ASTC</math>, <math>\angle SCT = \angle SAT</math>, so <math>\angle BXT = \angle SAT</math>. Since <math>ASTB</math> is cyclic, <math>\angle AST = \angle TBX</math>. Thus, <math>\triangle AST \sim \triangle XBT</math>, so <math>AS/XB = ST/BT</math>. Solving for <math>ST</math> yields <math>ST = \frac{35}{8}</math>, for a final answer of <math>35+8 = \boxed{043}</math>. | By the Inscribed Angle Theorem on <math>CPTX</math>, <math>\angle SCT = \angle BXT</math>. By the Inscribed Angle Theorem on <math>ASTC</math>, <math>\angle SCT = \angle SAT</math>, so <math>\angle BXT = \angle SAT</math>. Since <math>ASTB</math> is cyclic, <math>\angle AST = \angle TBX</math>. Thus, <math>\triangle AST \sim \triangle XBT</math>, so <math>AS/XB = ST/BT</math>. Solving for <math>ST</math> yields <math>ST = \frac{35}{8}</math>, for a final answer of <math>35+8 = \boxed{043}</math>. | ||
~ Leo.Euler | ~ Leo.Euler | ||
+ | |||
+ | ==Solution 5 (5 = 2 + 3)== | ||
+ | [[File:2016 AIME II 10.png|430px|right]] | ||
+ | By Ptolemy's Theorem applied to quadrilateral <math>ASTB</math>, we find | ||
+ | <cmath>AS\cdot BT+AB\cdot ST=AT\cdot BS.</cmath> | ||
+ | Projecting through <math>C</math> we have | ||
+ | <cmath>\frac{AQ \cdot PB}{PQ \cdot AB} = (A,Q; P,B)\stackrel{C}{=}(A,T; S,B)=\frac{AT \cdot BS}{ST \cdot AB}. </cmath> | ||
+ | Therefore <cmath>AT \cdot BS = \frac {AQ \cdot PB}{PQ} \times ST \implies</cmath> | ||
+ | <cmath>\left(\frac {AQ \cdot PB}{PQ} - AB\right)\times ST = AS \cdot BT \implies</cmath> | ||
+ | <cmath>ST = \frac {AS \cdot BT \cdot PQ}{AQ \cdot PB – AB \cdot PQ}</cmath> | ||
+ | <cmath>ST = \frac {7\cdot 5 \cdot 3}{7\cdot 9 – 13 \cdot 3 } = \frac {35}{8} \implies 35 + 8 = \boxed {43}.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | Connect <math>AT</math> and <math>\angle{SCT}=\angle{SAT}, \angle{ACS}=\angle{ATS}, \frac{ST}{\sin \angle{SAT}}=\frac{AS}{\sin \angle{ATS}}</math> | ||
+ | |||
+ | So we need to get the ratio of <math>\frac{\sin \angle{ACS}}{\sin \angle{SCT}}</math> | ||
+ | |||
+ | By clear observation <math>\triangle{CAQ}\sim \triangle{BTQ}</math>, we have <math>\frac{CQ}{AC}=\frac{6}{5}</math>, LOS tells <math>\frac{AC}{\sin \angle{CPA}}=\frac{4}{\sin \angle{ACS}}; \frac{CQ}{\sin \angle{CPQ}}=\frac{3}{\sin \angle{PCQ}}</math> so we get <math>\frac{\sin \angle{PCQ}}{\sin \angle{ACS}}=\frac{5}{8}</math>, the desired answer is <math>7\cdot \frac{\sin \angle{SAT}}{\sin \angle{ATS}}=\frac{35}{8}</math> leads to <math>\boxed{043}</math> | ||
+ | |||
+ | ~blusoul | ||
+ | |||
+ | ==Solution 7 (no trig or projections)== | ||
+ | |||
+ | Note that since <math>\triangle SAP~\triangle BCP</math>, <math>\frac{9}{SP}=\frac{BC}{7}=\frac{PC}{4}</math>. Furthermore, since <math>\triangle ACQ~\triangle TBQ</math>, we have <math>\frac{7}{TQ}=\frac{AC}{5}=\frac{QC}{6}</math>. From Stewart's on triangle <math>BCP</math>, we have <math>25CQ+BC^2\cdot TQ=TQ\cdot CQ\cdot TC+36TC</math>, and since <math>TQ\cdot CQ=6\cdot7=42</math> by power of a point, this simplifies to <math>25CQ+BC^2\cdot TQ=78TC</math>. Similarly, <math>49CP+AC^2\cdot SP=52SC</math>. Finally, using Ptolemy's on quadrilateral <math>ACBS</math> yields <math>13SC=7BC+SB\cdot AC</math>, and using Ptolemy's on quadrilateral <math>ACBT</math> yields <math>13TC=5AC+TA\cdot BC</math>. From Ptolemy's on <math>ABTS</math>, we find <math>SB\cdot TA=13ST+35</math>, which is nice because it contains <math>ST</math>. | ||
+ | We return to our first Stewart's equation: <math>25CQ+BC^2\cdot TQ=78TC</math>, and we notice that <math>CQ</math> and <math>TQ</math> can be related to <math>AC</math> using our similar triangle conditions. Substituting gives us <math>30AC+\frac{35BC^2}{AC}=78TC</math>, which by four times our first Ptolemy's equation also equals <math>30AC+6TA\cdot BC</math>. Thus, <math>\frac{35BC^2}{AC}=6TA\cdot BC</math> and <math>TA=\frac{35}{6}\cdot\frac{BC}{AC}</math>. Similarly, from our other Stewart's equation, we find <math>28BC+\frac{63AC^2}{BC}=52SC=28BC+4SB\cdot AC</math>, or <math>SB=\frac{63}{4}\cdot\frac{AC}{BC}</math>. Plugging this into our final Ptolemy's equation, we find <cmath>SB\cdot TA=13ST+35\Longrightarrow\frac{35\cdot63}{6\cdot4}=13ST+35\Longrightarrow ST=\frac{\frac{35\cdot21}{8}-35}{13}=\frac{35\cdot\frac{13}{8}}{13}=\frac{35}{8},</cmath>giving us our final answer of <math>\boxed{043}</math>. | ||
+ | |||
+ | ~wuwang2002 | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=9|num-a=11}} | {{AIME box|year=2016|n=II|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:49, 22 November 2024
Contents
Problem
Triangle is inscribed in circle . Points and are on side with . Rays and meet again at and (other than ), respectively. If and , then , where and are relatively prime positive integers. Find .
Solution 1
Let , , and . Note that since we have , so by the Ratio Lemma Similarly, we can deduce and hence .
Now Law of Sines on , , and yields Hence so Hence and the requested answer is .
Edit: Note that the finish is much simpler. Once you get , you can solve quickly from there getting .
Solution 2 (Projective Geometry)
Projecting through we have which easily gives .
Solution 3
By Ptolemy's Theorem applied to quadrilateral , we find Therefore, in order to find , it suffices to find . We do this using similar triangles, which can be found by using Power of a Point theorem.
As , we find Therefore, .
As , we find Therefore, .
As , we find Therefore, .
As , we find Therefore, . Thus we find But now we can substitute in our previously found values for and , finding Substituting this into our original expression from Ptolemy's Theorem, we find Thus the answer is .
Solution 4
Extend past to point so that is cyclic. Then, by Power of a Point on , . By Power of a Point on , . Thus, , so .
By the Inscribed Angle Theorem on , . By the Inscribed Angle Theorem on , , so . Since is cyclic, . Thus, , so . Solving for yields , for a final answer of .
~ Leo.Euler
Solution 5 (5 = 2 + 3)
By Ptolemy's Theorem applied to quadrilateral , we find Projecting through we have Therefore
vladimir.shelomovskii@gmail.com, vvsss
Solution 6
Connect and
So we need to get the ratio of
By clear observation , we have , LOS tells so we get , the desired answer is leads to
~blusoul
Solution 7 (no trig or projections)
Note that since , . Furthermore, since , we have . From Stewart's on triangle , we have , and since by power of a point, this simplifies to . Similarly, . Finally, using Ptolemy's on quadrilateral yields , and using Ptolemy's on quadrilateral yields . From Ptolemy's on , we find , which is nice because it contains . We return to our first Stewart's equation: , and we notice that and can be related to using our similar triangle conditions. Substituting gives us , which by four times our first Ptolemy's equation also equals . Thus, and . Similarly, from our other Stewart's equation, we find , or . Plugging this into our final Ptolemy's equation, we find giving us our final answer of .
~wuwang2002
See also
2016 AIME II (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.