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Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 3"

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''Virgil Nicula''
 
''Virgil Nicula''
 
==Solution==
 
==Solution==
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{{solution}}
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(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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{{image}}
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 +
 
 +
 
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(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>
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 +
And now we use the Law of Sines! On <math>\triangle BCD</math>!
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<cmath>CD=\frac{BC\sin{(\theta+30^{\circ})}}{\sin{(60^{\circ}-\frac{\theta}{2})}}</cmath>
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 +
Therefore
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<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>
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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
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<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>
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 +
This is a simple rearrangement of the Pythagorean Identity, which is true. We can work backwards to get that <math>PD=r</math>.
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==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 $ABCD$ inscribed in a circle of radius $r$, for which there is a point $P$ on $CD$ such that $CB=BP=PA=AB$.

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

(b) Prove that $PD=r$.

Virgil Nicula

Solution

(a) Note that, given the positions of $A$ and $B$, $P$ can be in exactly two places. However, $P$ is on segment $\overline{CD}$, and $C$ and $D$ are on the same side of line $\overleftrightarrow{AB}$, so $P$ must be on the same side of $\overline{AB}$ as $C$. This shows that, given the positions of $A$, $B$, and $C$, we can determine the position of $P$. Also note that $P$ must be in the circumcircle of $\triangle ABC$, which means that $C$ must be outside triangle $ABP$. Without loss of generality, assume that $C$ and $A$ are on opposite sides of $BP$. Now it suffices to show that for ever positioning of $A$, $B$, $C$, and $P$ such that $CB=BP=PA=AB$, there exists a $D$ on $\overline{CP}$ such that $ABCD$ is cyclic. This is simple; merely extend $\overline{CP}$ so that it intersects the circumcircle of $ABC$ again at $D$. An example of such an arrangement of points is shown below; $P=(0,0)$, $A=(-1,\sqrt{3})$, $B=(1,\sqrt{3})$, $C=(2,0)$, and $D=(-2,0)$.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


(b) Let $AB=BP=PA=BC=a$ and $\angle CBP=\theta$. I shall now find all of the angles of $\triangle BCD$. We know that $\angle ABP=60^{\circ}$, since $\triangle ABP$ is equilateral. Therefore $\angle ABC=60^{\circ}+\theta$. Triangle $ABC$ is isosceles, so we have that $AB=BC=a$, and $\angle BAC=60^{\circ}-\frac{\theta}{2}$. Since $\angle BAC$ and $\angle BDC$ are inscribed angles that intercept minor arc $\widehat{BC}$, $\angle BDC=\angle BAC = 60^{\circ}-\frac{\theta}{2}$. Now note that $\angle BCD=\angle BCP$. Since $\triangle BCP$ is isosceles, $\angle BCP=90^{\circ}-\frac{\theta}{2}$. We know that $\angle BDC=60^{\circ}-\frac{\theta}{2}$ and $\angle BCD=90^{\circ}-\frac{\theta}{2}$, so $\angle DBC=\theta+30^{\circ}$. We are now able to use the Law of Sines on $\triangle BCD$. However, we would only get an equation involving $CD$, so if we were to use it to find $PD$, we must find the length of $\overline{CP}$. This can easily be done using the Law of Sines on $\triangle BCP$:

\[CP=\frac{a\sin{\theta}}{\sin{(90^{\circ}-\frac{\theta}{2})}}\]

And now we use the Law of Sines! On $\triangle BCD$!

\[CD=\frac{BC\sin{(\theta+30^{\circ})}}{\sin{(60^{\circ}-\frac{\theta}{2})}}\]

Therefore

\[PD=CD-CP=\frac{a\sin{(\theta+30^{\circ})}}{\sin{(60^{\circ}-\frac{\theta}{2})}}-\frac{a\sin{\theta}}{\sin{(90^{\circ}-\frac{\theta}{2})}}\]

Using the Law of Sines on $\triangle BAC$ gives that $r=\frac{a}{2\sin{(60^{\circ}-\frac{\theta}{2})}}$, so it suffices to show that

\[\frac{a}{2\sin{(60^{\circ}-\frac{\theta}{2})}}=\frac{a\sin{(\theta+30^{\circ})}}{\sin{(60^{\circ}-\frac{\theta}{2})}}-\frac{a\sin{\theta}}{\sin{(90^{\circ}-\frac{\theta}{2})}}\]

Canceling out $a$'s and rearranging shows that this statement is equivalent to

\[\frac{2\sin{(\theta+30^{\circ})}-1}{2\sin{(60^{\circ}-\frac{\theta}{2})}}=\frac{\sin{\theta}}{\sin{(90^{\circ}-\frac{\theta}{2})}}\]

Using the sine and cosine addition formulae gives that this statement is equivalent to

\[\frac{\sqrt{3}\sin{\theta}+\cos{\theta}-1}{\sqrt{3}\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}}=\frac{\sin{\theta}}{\cos{\frac{\theta}{2}}}\]

Cross-multiplying and using double-angle formulae gives that this statement is equivalent to

\[2\sqrt{3}\sin{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}+\cos^3{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}\cos{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}=2\sqrt{3}\sin{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}-2\sin^2{\frac{\theta}{2}}\cos{\frac{\theta}{2}}\]

Canceling out like terms and dividing both sides by $\cos{\frac{\theta}{2}}$ gives that this statement is equivalent to

\[\cos^2{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}-1=-2\sin^2{\frac{\theta}{2}}\]

This is a simple rearrangement of the Pythagorean Identity, which is true. We can work backwards to get that $PD=r$.


See also

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