View source for 2006 Romanian NMO Problems/Grade 9/Problem 3

==Problem==
We have a [[quadrilateral]] <math>ABCD</math> [[inscribe]]d in a [[circle]] of [[radius]] <math>r</math>, for which there is a [[point]] <math>P</math> on <math>CD</math> such that <math>CB=BP=PA=AB</math>.

(a) Prove that there are points <math>A,B,C,D,P</math> which fulfill the above conditions.

(b) Prove that <math>PD=r</math>.

''Virgil Nicula''
==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>.

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Solution 2 (suli) Given the circle <math>O</math>, construct equilateral <math>ABP</math> inside <math>O</math> with <math>A</math>, <math>B</math> on its interior such that <math>AB < r</math>. Then let the circle centered at <math>B</math> with radius <math>BP</math> intersect circle <math>O</math> at <math>C</math>. Finally, extend <math>CP</math> to meet <math>O</math> at <math>D</math> and we are done!

(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>.


Soln. 2 (fmasroor)
Let <CBP=2x, so that <BPC=<BCP=90-x (no fractions). Then <ADC=180-<ABC=120-2x, <APD=180-<APB-<BPC=x+30, so then <PAD=x+30. Then PD=PA so se need to prove that ODA is equilateral where O is the center of ABCD. However, since <DAP=<DPA=x+30 DP=AP and so ABPD is a kite; <ABD=60/2=30 and then <DOA=2*30=60, so then ODA is equilateral. Done.

Solution 3 (suli)
Notice that the circumcenter of <math>APC</math> is <math>B</math>, so <math><ACP = \frac{<ABP}{2} = 30^\circ</math>. (Note that <math>ABP</math> is equilateral.) Hence <math><AOD = 2<ACD = 60^\circ</math>, so <math>ODA</math> is equilateral and <math>AD = r</math>. Now if we let <math><PBC = 2x</math> then it is easily seen via angle chasing that <math><DAP = <DPA = 30^\circ + x</math>, so <math>PD = PA = r</math> as desired.

==See also==
*[[2006 Romanian NMO Problems/Grade 9/Problem 2 | Previous problem]]
*[[2006 Romanian NMO Problems/Grade 9/Problem 4 | Next problem]]
*[[2006 Romanian NMO Problems]]
[[Category: Olympiad Geometry Problems]]