2006 Romanian NMO Problems/Grade 9/Problem 3

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 $D=(-2,0)$ .

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(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 $PD=r$ .

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.

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2006 Romanian NMO Problems/Grade 9/Problem 3

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