344. Some trigonometry problems.
by Virgil Nicula, Jun 18, 2012, 2:29 PM
PP0. Let 
and sequencies 
for 
; 
. Prove that 
. See and here.
PP1. Let
be a triangle so that 
and 
. Find the value of 
.
Proof 1 (only with angles). The relation means that exists 
so that 
. Thus, 
. Since obtain that 
.
Proof 2. From well-known property
obtain that 
.
Hence can suppose w.l.o.g. that
and 
.
Proof of the remarkable property. Denote 
for which 
. Thus, 
(the line
is tangent to the circumcircle of the triangle 
in the point 
) 
. Denote 
.
Proof 3.
.
PP1 bis. In any triangle exists the implication 
.
![\[\Longrightarrow\ \mathrm{(\ implication\ from\ left\ to\ right\ )}\]](//latex.artofproblemsolving.com/7/b/6/7b6ea8a62e098a44ae37068801e99c137f4c89b3.png)
Proof 1 (synthetic). Denote 
so that 
.
Thus,
, 
and 
and 
.
.
Proof 2 (metric). Denote
so that 
. Thus 
, 
, 
,
. Apply the Srewart's relation to the ray 
in 
.
Proof 3 (trigonometric).
![$c=a[3-2(1-\cos 2A)]\implies$](//latex.artofproblemsolving.com/5/0/c/50cbd1dba9c3a355610e512e94d8e73a78202c23.png)
. Thus, 
.
Proof 4. Denote so that . Thus, , 
and , 
.
Apply the well-known relation
.
![\[\Longleftrightarrow\ \mathrm{(\ equivalence\ )}\]](//latex.artofproblemsolving.com/2/b/0/2b0e29c93ebf868847760a963774f7d5e48f332e.png)
Proof. 
. In conclusion. .
PP2. Let
be the length-sides of ,so that 
, 
and 
. Find the value of 
.
Proof.
, i.e. 
. Apply Pythagoras' relation 
. Remark that 
.
PP3. Prove that
.
Proof 1. Denote
. The assumption becomes 
, i.e. 
and the conclusion becomes 
![$\left[(a+b)u^2+(a-b)\right]\cdot\left[(a+b)v^2+(a-b)\right]=\left(a^2-b^2\right)\left(1+u^2\right)\left(1+v^2\right)\iff$](//latex.artofproblemsolving.com/f/0/a/f0a4abcc53357e2f60d983c8bd7f019c68669120.png)
![$\left(a^2-b^2\right)\left[1+\left(u^2+v^2\right)+u^2v^2\right]$](//latex.artofproblemsolving.com/a/2/c/a2c6a1a04aa4bf3dca537ed8daae0201f10ddf8d.png)
, what is truly because 
and 
.
Proof 2. 
.
PP4. Let be the length-sides of a triangle 
, where 
. Find the value of .
Proof. Prove easily that
, where . From the relations 
obtain
Apply the relation
.
Remark. Prove easily that
. Thus, 
.
PP5. Let with 
. Find the area ![$S=[ABC]$](//latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
.
Proof 1.
and 
. Since 
obtain 
.
From the relations
and 
get 
, i.e. 
. In conclusion, 
.
Proof 2 (GANITEE).
and 
. Using the well-known identity
obtain that 
. In conclusion,
.
An easy extension. Let be a triangle with 
. Then 
.
Proof.
Since
obtain that 
. Therefore, 
because .
.
Remark. The proposed problem is a particular case of this characterization because
, what is truly. Thus,
, 
and denote 
, where 
is the midpoint of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Then 
, i.e. 
and 
.
PP6. An isosceles triangle has its orthocentre lying on its incircle. Prove that
.
Proof. Let be an 
-isosceles triangle with the incircle 
and the circumcircle 
. Denote midpoint of . Thus,
.
.
Remark. I used the remarkable identity
. In the particular
case
this identity becomes 
.
PP7. An-isosceles triangle has its circumcentre lying on its incircle. Prove that 
.
Proof.
and 
.
PP8. An-isosceles triangle has its Nagel point lying on its incircle. Prove that 
.
Proof. Denote
, where 
is the Nagel point of , i.e. 
and 
.
Apply the Menelaus' theorem to the transversal
for 
. Therefore, 
and 
.
PP9. In a triangle , 
and 
. Find the value 
. (China JiangXi 2012)
Proof.
. I used the well-known identity 
.
PP10. Let with the circymcircle 
. Let 
be the midpoint of and 
such that 
, where 
. Prove that .
Method 1.
, where 
.
Method 2. Denote
, where 
is the tangent in to 
. Prove easily that 
.
PP11. Solve the equation
.
Proof.
.
PP12. Solve the equation
.
Proof.
.
PP13. Solve the equation
, where 
.
Proof.
and 
a.s.o.
PP14. Let
be a rhombus.The circumradius of 
is 
and the circumradius of 
is 
. Find the area of rhombus.
Proof. Let
and 
. Thus, 
and ![$[ABCD]=a^2\sin\phi=$](//latex.artofproblemsolving.com/8/4/a/84a9101fa55bff72c17eb2b822e10ee708140d00.png)
![$\boxed{\ [ABCD]=\frac {8R^3r^3}{\left(R^2+r^2\right)^2}\le 2Rr\ }$](//latex.artofproblemsolving.com/0/2/1/0216a165d0a78cebdf4c155edff3f52b16a3f6c9.png)
.
PP15. Letand 
such that 
. Find 
.
Proof 1 (trigonometric).
.
Proof 2 (trigonometric).
.
Proof 3 (trigonometric). Let
so that 
and 
. Thus, 
.
PP16.
is an equilateral triangle.
Proof 1.![\[\frac{\sin B\cos C}{\sin A} =\frac{\sin C\cos A}{\sin B} = \frac{\sin A\cos B}{\sin C}\]](//latex.artofproblemsolving.com/c/1/1/c1114ceed879704b75f87a4749d33a9e63863c40.png)
![\[\frac ba\cdot\frac {a^2+b^2-c^2}{2ab} =\frac cb\cdot\frac {b^2+c^2-a^2}{2bc} =\frac ac\cdot\frac {c^2+a^2-b^2}{2ca}\]](//latex.artofproblemsolving.com/a/1/6/a164d10c1b0f268d4ac7f040b69c7fd4f7694825.png)
![\[\frac {b^2-c^2}{a^2}=\frac {c^2-a^2}{b^2}=\frac {a^2-b^2}{c^2}=\frac {\left(b^2-c^2\right)+\left(c^2-a^2\right)+\left(a^2-b^2\right)}{a^2+b^2+c^2}\]](//latex.artofproblemsolving.com/e/e/7/ee712ca74bfaaa269da93dd929175271232fae4c.png)
![\[a=b=c\]](//latex.artofproblemsolving.com/4/8/8/4882ed810e2fa587feccdd728930ecd4f8708885.png)
Proof 2. ![\[\frac {\sin (B-C)}{\sin (B+C)} =\frac {\sin (C-A)}{\sin (C+A)} =\frac {\sin (A-B)}{\sin (A+B)}\]](//latex.artofproblemsolving.com/d/b/1/db1159c2b64d417ba1f49333570b64b06f4f8d77.png)
![\[\frac {\sin B\cos C}{\sin C\cos B}=\frac {\sin C\cos A}{\sin A\cos C}=\frac {\sin A\cos B}{\sin B\cos A}\]](//latex.artofproblemsolving.com/7/4/5/745d198af18d215a3a8b3e58f9413118047578fc.png)
![\[\frac {\tan B}{\tan C}=\frac {\tan C}{\tan A}=\frac {\tan A}{\tan B}=\sqrt [3]{\frac {\tan B}{\tan C}\cdot\frac {\tan C}{\tan A}\cdot\frac {\tan A}{\tan B}}=1\]](//latex.artofproblemsolving.com/4/0/0/4007258c42648e1773ec74802b3ac937e993cf5e.png)
![\[\tan A=\tan B=\tan C\]](//latex.artofproblemsolving.com/5/d/f/5df455366553b066ed86876935ebe179219da536.png)
PP17. Let
and the midpoint of ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
such that 
and 
. Prove that 
.
Proof.
and 
. I used the well-known property 
.
PP18. Let be an -isosceles with the ortocenter 
and the incircle 
. Prove that 
.
Proof. Let the midpoint of and suppose w.l.o.g. 
. Thus, 
, 
. Thus,
.
Otherwise.
. I used the identities 
and 
.
Remark. In any triangle there is the equivalence 
.
If , then becomes 
a.s.o.
PP19. If in a triangle
, median 
is such that 
and 
, then prove that 
.
Proof 1 (synthetic). Denote
and the second intersection 
of 
with the circumcircle of .
Prove easily that
. In conclusion, .
Proof 2.
. Apply an well-known property 
. Thus, 
![$\frac {\sin \left[3\left(45^{\circ}-C\right)\right]}{\sin\left(45^{\circ}-C\right)}=$](//latex.artofproblemsolving.com/5/7/e/57e79ab34a5b42507c232572b729890e1564466c.png)
.
PP20. Prove that
.
Proof.
![$\frac 12\cdot\sum_{\mathrm{cyc}}\left[\cos (2\gamma +2\theta )-\cos (2\beta +2\theta )\right]=0$](//latex.artofproblemsolving.com/4/5/3/453306633282b3ce13f1c4d3bccfb30ba82c8900.png)
.
PP21. Find
such that 
.
Proof.
, i.e. 
![$-\frac {\sin\left[3\left(x-\frac {\pi}{4}\right)\right]}{\sin\left(x-\frac {\pi}{4}\right)}=$](//latex.artofproblemsolving.com/6/c/8/6c8b2e63609594bd6c63c307e123cca7146377e9.png)
![$2\left[1-\cos 2\left(x-\frac {\pi}{4}\right)\right]-3=-1-2\sin 2x$](//latex.artofproblemsolving.com/b/b/f/bbf79188828cf18e584a63f623b7a368dd0b69ea.png)
. Therefore, 
. We can use
the standard notation
, where 
is the period. In the our case, 
.
PP22. Solve the trigonometrical equation
.
Proof.
.
PP23. Let be a triangle for which 
and 
. Find the value of .
Proof 1 (with lengths).
, Therefore, 
. These values correspond to a 
triangle.
Proof 2 (with angles). Firstly, by using the law of sines in the given equality i.e. we obtain 
and since 
it means that 
.
On the other hand, by the law of cosines one has:
which further rewrites as 
, where at the last step I used again the law of sines. But taking into account the above derived relation , the latter equality reduces
to:
. Now denote 
to rewrite the previous relation as a quadratic equation: 
, which has the solutions 
.
Hence
and thus 
but since 
it means that the only possible values for are 
or 
which leads to: 
.
Now it's easy to check that these values fullfill the given conditions, as they correspond to a triangle.
PP24. Solve the equation
.
Proof. The relation 
.
PP25. Let be a triangle for which 
. What type of triangle is this ?
Proof 1 (trig. - vanstraelen).
.
Proof 2 (metric - Faustus). 
![$a^2\cdot \left[(b+c)^2-a^2\right]=(b+c)^2\cdot\left[a^2-(b-c)^2\right]\iff$](//latex.artofproblemsolving.com/a/6/a/a6a5a8efc705ea27a79bc26177b5f5811b0ab0f2.png)
.
Proof 3 (algebraic - Farenhajt). Denote
, where 
and 
. Thus,
.
PP26. Let
be a rectangle with 
. Let 
so that 
and 
. Show that 
.
Proof 1 (trigonometric). Suppose w.l.o.g. that
. Thus, 
and 
, 
and 
.
In conclusion,
.
Proof 2 (trigonometric). Denote
and 
such that 
and 
. Prove easily that 
. Obtain that
is 
-rightangled and isosceles triangle and 
.
PP27. Let be a square with 
and 
, 
so that 
and 
. Find 
.
Proof 1. Denote
so that 
and 
. Observe that
and 
.
Proof 2. Denote
. Observe that 
and
.
Remark.
.
PP28. Let be a triangle so that 
. Prove that 
.
Proof.
, 
, 
for some positive real number 
. Using the well-known identity 
obtain
that
and 
, 
,
, 
, 
. Therefore, .
PP29. Prove that in there is the implication 
.
Proof 1. Take regular
with 
. The Ptolemy's relation in 
.
Proof 2.
. Since 
obtain 
So
. From 
obtain that 
. But 
- truly.
Proof 3. Let
. So 
. Denote
. Thus, 
. Thus 
. Denote 
. Thus, 
, 
and
.
and sequencies 
for 
; 
. Prove that 
. See and here.PP1. Let
be a triangle so that 
and 
. Find the value of 
.Proof 1 (only with angles). The relation
means that exists 
so that 
. Thus, 
. Since obtain that 
.Proof 2. From well-known property
obtain that 
.Hence can suppose w.l.o.g. that
and 
.
Proof of the remarkable property. Denote 
for which 
. Thus, 
(the line
is tangent to the circumcircle of the triangle 
in the point 
) 
. Denote 
.Proof 3.
.PP1 bis. In any triangle
exists the implication 
.![\[\Longrightarrow\ \mathrm{(\ implication\ from\ left\ to\ right\ )}\]](http://latex.artofproblemsolving.com/7/b/6/7b6ea8a62e098a44ae37068801e99c137f4c89b3.png)
Proof 1 (synthetic). Denote 
so that 
.Thus,
, 
and 
and 
.
.Proof 2 (metric). Denote
so that 
. Thus 
, 
, 
, . Apply the Srewart's relation to the ray 
in 
.Proof 3 (trigonometric).
![$c=a[3-2(1-\cos 2A)]\implies$](http://latex.artofproblemsolving.com/5/0/c/50cbd1dba9c3a355610e512e94d8e73a78202c23.png)
. Thus, 
.Proof 4. Denote
so that . Thus, , 
and , 
.Apply the well-known relation
.![\[\Longleftrightarrow\ \mathrm{(\ equivalence\ )}\]](http://latex.artofproblemsolving.com/2/b/0/2b0e29c93ebf868847760a963774f7d5e48f332e.png)
Proof. 
. In conclusion. .PP2. Let
be the length-sides of ,so that 
, 
and 
. Find the value of 
.Proof.
, i.e. 
. Apply Pythagoras' relation 
. Remark that 
.PP3. Prove that
.Proof 1. Denote
. The assumption becomes 
, i.e. 
and the conclusion becomes 
![$\left[(a+b)u^2+(a-b)\right]\cdot\left[(a+b)v^2+(a-b)\right]=\left(a^2-b^2\right)\left(1+u^2\right)\left(1+v^2\right)\iff$](http://latex.artofproblemsolving.com/f/0/a/f0a4abcc53357e2f60d983c8bd7f019c68669120.png)
![$\left(a^2-b^2\right)\left[1+\left(u^2+v^2\right)+u^2v^2\right]$](http://latex.artofproblemsolving.com/a/2/c/a2c6a1a04aa4bf3dca537ed8daae0201f10ddf8d.png)
, what is truly because 
and 
.Proof 2.
.PP4. Let
be the length-sides of a triangle 
, where 
. Find the value of .Proof. Prove easily that
, where . From the relations 
obtain
Apply the relation
.Remark. Prove easily that
. Thus, 
.PP5. Let
with 
. Find the area ![$S=[ABC]$](http://latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
.Proof 1.
and 
. Since 
obtain 
.From the relations
and 
get 
, i.e. 
. In conclusion, 
.Proof 2 (GANITEE).
and 
. Using the well-known identity
obtain that 
. In conclusion,
.An easy extension. Let
be a triangle with 
. Then 
. Proof.
Since
obtain that 
. Therefore, 
because .
.Remark. The proposed problem is a particular case of this characterization because
, what is truly. Thus, , 
and denote 
, where 
is the midpoint of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Then 
, i.e. 
and 
.PP6. An isosceles triangle has its orthocentre lying on its incircle. Prove that
.Proof. Let
be an 
-isosceles triangle with the incircle 
and the circumcircle 
. Denote midpoint of . Thus,
.
.Remark. I used the remarkable identity
. In the particularcase
this identity becomes 
.PP7. An
-isosceles triangle has its circumcentre lying on its incircle. Prove that 
.Proof.
and 
.PP8. An
-isosceles triangle has its Nagel point lying on its incircle. Prove that 
.Proof. Denote
, where 
is the Nagel point of , i.e. 
and 
.Apply the Menelaus' theorem to the transversal
for 
. Therefore, 
and 
.PP9. In a triangle
, 
and 
. Find the value 
. (China JiangXi 2012)Proof.
. I used the well-known identity 
.PP10. Let
with the circymcircle 
. Let 
be the midpoint of and 
such that 
, where 
. Prove that .Method 1.
, where 
.Method 2. Denote
, where 
is the tangent in to 
. Prove easily that 
.PP11. Solve the equation
.Proof.
.PP12. Solve the equation
.Proof.
.PP13. Solve the equation
, where 
.Proof.
and 
a.s.o.PP14. Let
be a rhombus.The circumradius of 
is 
and the circumradius of 
is 
. Find the area of rhombus.Proof. Let
and 
. Thus, 
and ![$[ABCD]=a^2\sin\phi=$](http://latex.artofproblemsolving.com/8/4/a/84a9101fa55bff72c17eb2b822e10ee708140d00.png)
![$\boxed{\ [ABCD]=\frac {8R^3r^3}{\left(R^2+r^2\right)^2}\le 2Rr\ }$](http://latex.artofproblemsolving.com/0/2/1/0216a165d0a78cebdf4c155edff3f52b16a3f6c9.png)
.PP15. Let
and 
such that 
. Find 
.Proof 1 (trigonometric).
.Proof 2 (trigonometric).
.Proof 3 (trigonometric). Let
so that 
and 
. Thus, 
.PP16.
is an equilateral triangle.Proof 1.
![\[\frac{\sin B\cos C}{\sin A} =\frac{\sin C\cos A}{\sin B} = \frac{\sin A\cos B}{\sin C}\]](http://latex.artofproblemsolving.com/c/1/1/c1114ceed879704b75f87a4749d33a9e63863c40.png)
![\[\frac ba\cdot\frac {a^2+b^2-c^2}{2ab} =\frac cb\cdot\frac {b^2+c^2-a^2}{2bc} =\frac ac\cdot\frac {c^2+a^2-b^2}{2ca}\]](http://latex.artofproblemsolving.com/a/1/6/a164d10c1b0f268d4ac7f040b69c7fd4f7694825.png)
![\[\frac {b^2-c^2}{a^2}=\frac {c^2-a^2}{b^2}=\frac {a^2-b^2}{c^2}=\frac {\left(b^2-c^2\right)+\left(c^2-a^2\right)+\left(a^2-b^2\right)}{a^2+b^2+c^2}\]](http://latex.artofproblemsolving.com/e/e/7/ee712ca74bfaaa269da93dd929175271232fae4c.png)
![\[a=b=c\]](http://latex.artofproblemsolving.com/4/8/8/4882ed810e2fa587feccdd728930ecd4f8708885.png)
Proof 2. ![\[\frac {\sin (B-C)}{\sin (B+C)} =\frac {\sin (C-A)}{\sin (C+A)} =\frac {\sin (A-B)}{\sin (A+B)}\]](http://latex.artofproblemsolving.com/d/b/1/db1159c2b64d417ba1f49333570b64b06f4f8d77.png)
![\[\frac {\sin B\cos C}{\sin C\cos B}=\frac {\sin C\cos A}{\sin A\cos C}=\frac {\sin A\cos B}{\sin B\cos A}\]](http://latex.artofproblemsolving.com/7/4/5/745d198af18d215a3a8b3e58f9413118047578fc.png)
![\[\frac {\tan B}{\tan C}=\frac {\tan C}{\tan A}=\frac {\tan A}{\tan B}=\sqrt [3]{\frac {\tan B}{\tan C}\cdot\frac {\tan C}{\tan A}\cdot\frac {\tan A}{\tan B}}=1\]](http://latex.artofproblemsolving.com/4/0/0/4007258c42648e1773ec74802b3ac937e993cf5e.png)
![\[\tan A=\tan B=\tan C\]](http://latex.artofproblemsolving.com/5/d/f/5df455366553b066ed86876935ebe179219da536.png)
PP17. Let
and the midpoint of ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
such that 
and 
. Prove that 
.Proof.
and 
. I used the well-known property 
.PP18. Let
be an -isosceles with the ortocenter 
and the incircle 
. Prove that 
.Proof. Let the midpoint
of and suppose w.l.o.g. 
. Thus, 
, 
. Thus,
.Otherwise.
. I used the identities 
and 
.Remark. In any triangle
there is the equivalence 
.If
, then becomes 
a.s.o.PP19. If in a triangle
, median 
is such that 
and 
, then prove that 
.Proof 1 (synthetic). Denote
and the second intersection 
of 
with the circumcircle of .Prove easily that
. In conclusion, .Proof 2.
. Apply an well-known property 
. Thus, 
![$\frac {\sin \left[3\left(45^{\circ}-C\right)\right]}{\sin\left(45^{\circ}-C\right)}=$](http://latex.artofproblemsolving.com/5/7/e/57e79ab34a5b42507c232572b729890e1564466c.png)
.PP20. Prove that
. Proof.
![$\frac 12\cdot\sum_{\mathrm{cyc}}\left[\cos (2\gamma +2\theta )-\cos (2\beta +2\theta )\right]=0$](http://latex.artofproblemsolving.com/4/5/3/453306633282b3ce13f1c4d3bccfb30ba82c8900.png)
.PP21. Find
such that 
.Proof.
, i.e. 
![$-\frac {\sin\left[3\left(x-\frac {\pi}{4}\right)\right]}{\sin\left(x-\frac {\pi}{4}\right)}=$](http://latex.artofproblemsolving.com/6/c/8/6c8b2e63609594bd6c63c307e123cca7146377e9.png)
![$2\left[1-\cos 2\left(x-\frac {\pi}{4}\right)\right]-3=-1-2\sin 2x$](http://latex.artofproblemsolving.com/b/b/f/bbf79188828cf18e584a63f623b7a368dd0b69ea.png)
. Therefore, 
. We can usethe standard notation
, where 
is the period. In the our case, 
.PP22. Solve the trigonometrical equation
.Proof.
.PP23. Let
be a triangle for which 
and 
. Find the value of .Proof 1 (with lengths).
, Therefore, 
. These values correspond to a 
triangle.Proof 2 (with angles). Firstly, by using the law of sines in the given equality i.e.
we obtain 
and since 
it means that 
.On the other hand, by the law of cosines one has:
which further rewrites as 
, where at the last step I used again the law of sines. But taking into account the above derived relation , the latter equality reducesto:
. Now denote 
to rewrite the previous relation as a quadratic equation: 
, which has the solutions 
.Hence
and thus 
but since 
it means that the only possible values for are 
or 
which leads to: 
.Now it's easy to check that these values fullfill the given conditions, as they correspond to a
triangle.PP24. Solve the equation
.Proof. The relation
.PP25. Let
be a triangle for which 
. What type of triangle is this ?Proof 1 (trig. - vanstraelen).
.Proof 2 (metric - Faustus).
![$a^2\cdot \left[(b+c)^2-a^2\right]=(b+c)^2\cdot\left[a^2-(b-c)^2\right]\iff$](http://latex.artofproblemsolving.com/a/6/a/a6a5a8efc705ea27a79bc26177b5f5811b0ab0f2.png)
.Proof 3 (algebraic - Farenhajt). Denote
, where 
and 
. Thus, 
.PP26. Let
be a rectangle with 
. Let 
so that 
and 
. Show that 
.Proof 1 (trigonometric). Suppose w.l.o.g. that
. Thus, 
and 
, 
and 
.In conclusion,
.Proof 2 (trigonometric). Denote
and 
such that 
and 
. Prove easily that 
. Obtain that
is 
-rightangled and isosceles triangle and 
.PP27. Let
be a square with 
and 
, 
so that 
and 
. Find 
.Proof 1. Denote
so that 
and 
. Observe that
and 
.Proof 2. Denote
. Observe that 
and
.Remark.
.PP28. Let
be a triangle so that 
. Prove that 
.Proof.
, 
, 
for some positive real number 
. Using the well-known identity 
obtainthat
and 
, 
,
, 
, 
. Therefore, .PP29. Prove that in
there is the implication 
.Proof 1. Take regular
with 
. The Ptolemy's relation in 
.Proof 2.
. Since 
obtain 
So
. From 
obtain that 
. But 
- truly.Proof 3. Let
. So 
. Denote
. Thus, 
. Thus 
. Denote 
. Thus, 
, 
and
.This post has been edited 83 times. Last edited by Virgil Nicula, Nov 26, 2015, 12:42 PM
August 2018
January 2018