Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 3"
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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>: | (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>: | ||
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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>. | 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== | ==See also== |
Latest revision as of 19:28, 21 September 2014
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.
Solution 2 (suli) Given the circle , construct equilateral
inside
with
,
on its interior such that
. Then let the circle centered at
with radius
intersect circle
at
. Finally, extend
to meet
at
and we are done!
(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 .
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 is
, so
. (Note that
is equilateral.) Hence
, so
is equilateral and
. Now if we let
then it is easily seen via angle chasing that
, so
as desired.