2006 Romanian NMO Problems/Grade 9/Problem 3

Revision as of 18:28, 21 September 2014 by Suli (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.


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

(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.

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

See also

Invalid username
Login to AoPS