Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 3"
m |
|||
Line 8: | Line 8: | ||
''Virgil Nicula'' | ''Virgil Nicula'' | ||
==Solution== | ==Solution== | ||
− | {{ | + | (a) Note that, given the positions of <math>A</math> and <math>B</math>, <math>P</math> can be in exactly two places. However, <math>P</math> is on segment <math>\overline{CD}</math>, and <math>C</math> and <math>D</math> are on the same side of line <math>\overleftrightarrow{AB}</math>, so <math>P</math> must be on the same side of <math>\overline{AB}</math> as <math>C</math>. This shows that, given the positions of <math>A</math>, <math>B</math>, and <math>C</math>, we can determine the position of <math>P</math>. Also note that <math>P</math> must be in the circumcircle of <math>\triangle ABC</math>, which means that <math>C</math> must be outside triangle <math>ABP</math>. Without loss of generality, assume that <math>C</math> and <math>A</math> are on opposite sides of <math>BP</math>. Now it suffices to show that for ever positioning of <math>A</math>, <math>B</math>, <math>C</math>, and <math>P</math> such that <math>CB=BP=PA=AB</math>, there exists a <math>D</math> on <math>\overline{CP}</math> such that <math>ABCD</math> is cyclic. This is simple; merely extend <math>\overline{CP}</math> so that it intersects the circumcircle of <math>ABC</math> again at <math>D</math>. An example of such an arrangement of points is shown below; <math>P=(0,0)</math>, <math>A=(-1,\sqrt{3})</math>, <math>B=(1,\sqrt{3})</math>, <math>C=(2,0)</math>, and <math>D=(-2,0)</math>. |
+ | |||
+ | {{image}} | ||
+ | |||
+ | |||
+ | |||
+ | (b) Let <math>AB=BP=PA=BC=a</math> and <math>\angle CBP=\theta</math>. I shall now find all of the angles of <math>\triangle BCD</math>. We know that <math>\angle ABP=60^{\circ}</math>, since <math>\triangle ABP</math> is equilateral. Therefore <math>\angle ABC=60^{\circ}+\theta</math>. Triangle <math>ABC</math> is isosceles, so we have that <math>AB=BC=a</math>, and <math>\angle BAC=60^{\circ}-\frac{\theta}{2}</math>. Since <math>\angle BAC</math> and <math>\angle BDC</math> are inscribed angles that intercept minor arc <math>\widehat{BC}</math>, <math>\angle BDC=\angle BAC = 60^{\circ}-\frac{\theta}{2}</math>. Now note that <math>\angle BCD=\angle BCP</math>. Since <math>\triangle BCP</math> is isosceles, <math>\angle BCP=90^{\circ}-\frac{\theta}{2}</math>. We know that <math>\angle BDC=60^{\circ}-\frac{\theta}{2}</math> and <math>\angle BCD=90^{\circ}-\frac{\theta}{2}</math>, so <math>\angle DBC=\theta+30^{\circ}</math>. We are now able to use the Law of Sines on <math>\triangle BCD</math>. However, we would only get an equation involving <math>CD</math>, so if we were to use it to find <math>PD</math>, we must find the length of <math>\overline{CP}</math>. This can easily be done using the Law of Sines on <math>\triangle BCP</math>: | ||
+ | |||
+ | <cmath>CP=\frac{a\sin{\theta}}{\sin{(90^{\circ}-\frac{\theta}{2})}}</cmath> | ||
+ | |||
+ | And now we use the Law of Sines! On <math>\triangle BCD</math>! | ||
+ | |||
+ | <cmath>CD=\frac{BC\sin{(\theta+30^{\circ})}}{\sin{(60^{\circ}-\frac{\theta}{2})}}</cmath> | ||
+ | |||
+ | Therefore | ||
+ | |||
+ | <cmath>PD=CD-CP=\frac{a\sin{(\theta+30^{\circ})}}{\sin{(60^{\circ}-\frac{\theta}{2})}}-\frac{a\sin{\theta}}{\sin{(90^{\circ}-\frac{\theta}{2})}}</cmath> | ||
+ | |||
+ | Using the Law of Sines on <math>\triangle BAC</math> gives that <math>r=\frac{a}{2\sin{(60^{\circ}-\frac{\theta}{2})}}</math>, so it suffices to show that | ||
+ | |||
+ | <cmath>\frac{a}{2\sin{(60^{\circ}-\frac{\theta}{2})}}=\frac{a\sin{(\theta+30^{\circ})}}{\sin{(60^{\circ}-\frac{\theta}{2})}}-\frac{a\sin{\theta}}{\sin{(90^{\circ}-\frac{\theta}{2})}}</cmath> | ||
+ | |||
+ | Canceling out <math>a</math>'s and rearranging shows that this statement is equivalent to | ||
+ | |||
+ | <cmath>\frac{2\sin{(\theta+30^{\circ})}-1}{2\sin{(60^{\circ}-\frac{\theta}{2})}}=\frac{\sin{\theta}}{\sin{(90^{\circ}-\frac{\theta}{2})}}</cmath> | ||
+ | |||
+ | Using the sine and cosine addition formulae gives that this statement is equivalent to | ||
+ | |||
+ | <cmath>\frac{\sqrt{3}\sin{\theta}+\cos{\theta}-1}{\sqrt{3}\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}}=\frac{\sin{\theta}}{\cos{\frac{\theta}{2}}}</cmath> | ||
+ | |||
+ | Cross-multiplying and using double-angle formulae gives that this statement is equivalent to | ||
+ | |||
+ | <cmath>2\sqrt{3}\sin{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}+\cos^3{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}\cos{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}=2\sqrt{3}\sin{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}-2\sin^2{\frac{\theta}{2}}\cos{\frac{\theta}{2}}</cmath> | ||
+ | |||
+ | Canceling out like terms and dividing both sides by <math>\cos{\frac{\theta}{2}}</math> gives that this statement is equivalent to | ||
+ | |||
+ | <cmath>\cos^2{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}-1=-2\sin^2{\frac{\theta}{2}}</cmath> | ||
+ | |||
+ | This is a simple rearrangement of the Pythagorean Identity, which is true. We can work backwards to get that <math>PD=r</math>. | ||
+ | |||
+ | |||
==See also== | ==See also== | ||
*[[2006 Romanian NMO Problems/Grade 9/Problem 2 | Previous problem]] | *[[2006 Romanian NMO Problems/Grade 9/Problem 2 | Previous problem]] |
Revision as of 14:37, 11 December 2011
Problem
We have a quadrilateral inscribed in a circle of radius , for which there is a point on such that .
(a) Prove that there are points which fulfill the above conditions.
(b) Prove that .
Virgil Nicula
Solution
(a) Note that, given the positions of and , can be in exactly two places. However, is on segment , and and are on the same side of line , so must be on the same side of as . This shows that, given the positions of , , and , we can determine the position of . Also note that must be in the circumcircle of , which means that must be outside triangle . Without loss of generality, assume that and are on opposite sides of . Now it suffices to show that for ever positioning of , , , and such that , there exists a on such that is cyclic. This is simple; merely extend so that it intersects the circumcircle of again at . An example of such an arrangement of points is shown below; , , , , and .
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
(b) Let and . I shall now find all of the angles of . We know that , since is equilateral. Therefore . Triangle is isosceles, so we have that , and . Since and are inscribed angles that intercept minor arc , . Now note that . Since is isosceles, . We know that and , so . We are now able to use the Law of Sines on . However, we would only get an equation involving , so if we were to use it to find , we must find the length of . This can easily be done using the Law of Sines on :
And now we use the Law of Sines! On !
Therefore
Using the Law of Sines on gives that , so it suffices to show that
Canceling out 's and rearranging shows that this statement is equivalent to
Using the sine and cosine addition formulae gives that this statement is equivalent to
Cross-multiplying and using double-angle formulae gives that this statement is equivalent to
Canceling out like terms and dividing both sides by gives that this statement is equivalent to
This is a simple rearrangement of the Pythagorean Identity, which is true. We can work backwards to get that .