422. Geometrical problems.
by Virgil Nicula, Mar 28, 2015, 11:47 PM
PP1 (Cristian Tello). Let an 
-right 
with the -altitude 
, where 
and 
. Denote 
the incircle 
of
and the incircle 
of 
the tangent line 
to 
at 
and the tangent line 
to 
at 
, where
the incircle 
of 
and the incircle 
of 
Prove that 
Proof. Let
. Thus, 
and 
PP2 (Ruben Dario). Let-right with 
and altitude , where . Prove that 
so that 
with 
and in this
case, if denote
and 
, then 
and 
are the bisectors of 
and 
.
Proof 1. Apply the Ceva's theorem 
. Apply the theorem of cathetus
The difference of relations
and 
, i.e. 
. Exists at least one 
so that 
In conclusion, there is only one positive root 
, i.e. there is only one 
which respects mentioned
conditions. Thus,
and 
, i.e. the ray 
is the bisector of
Proof 2. Apply the Ceva's theorem
. Apply the theorem of the symmedian 
Hence 
a.s.o.
Remark.
the ray is the bisector of . Let 
so that
and 
. Prove easily that 
and 
. Hence 
and 
.
An easy extension. Let with -symmedian , where . Suppose that exists so that 
Let
. Prove that 
, 
and 
PP3 (M. Ochoa Sanchez). Let an acute with orthocenter 
circumcircle 
and 
Prove that 
Proof. Suppose w.l.o.g.
and let 
. Prove easily 
. Apply the theorem of Sines
in
the relation 
.
Remark.
is equilateral.
PP4 (M. Ochoa Sanchez). Let with 
and so that 
. Prove that 
.
Proof. Denote
. Apply the theorem of Sines in the triangles 
.
PP5. Let with 
. Prove that 
Proof. Suppose w.l.o.g.
. Let 
and 
. Is well-known that 
and 
. Therefore
.
.
L1.
there is the equivalence 
(standard notations).
Proof. Observe that
a.s.o. Thus, 
.
Remark 1. Denote the midpoint
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the incircle 
and the -excircle 
. Is well-known that
is the midpoint of
, the midpoint 
(south) of ![$[II_a]$](//latex.artofproblemsolving.com/b/9/d/b9d69f173ce4bad72631095440791204d2b021e3.png)
belongs to the circumcircle of and from the trapezoid 
obtain that 
,
where
is the length of the arrow of 
. Prove easily that the Ruben Dario's equivalence 
.
Remark 2.
![$\frac 14\cdot\left[\frac {s^2}{(s-b)(s-c)}-\frac {as}{(s-b)(s-c)}+1\right]=$](//latex.artofproblemsolving.com/a/e/9/ae9b6b3aacf77f776cd31605c2eda916ba83a253.png)
![$\frac 14\cdot\left[1+\frac {s(s-a)}{(s-b)(s-c)}\right]=$](//latex.artofproblemsolving.com/3/5/3/353f474e7db769c22aa63e39cfa43237187171b6.png)
.
PP6 (Ruben Dario). Let with the incircle 
which touches at 
, 
and 
. Let 
. Prove that 
.
Proof. Prove easily or is well-known that
and 
. Thus, the points 
belong to the circle with the dameter ![$[AI]$](//latex.artofproblemsolving.com/c/0/8/c08892aba53b466068fea586787c79e1aa04654a.png)
.
In conclusion,
. Nice problem!
P7 (Ruben Dario). Let with 
. Prove that 
.
Proof 1. Apply Menelaus' theorem to
.
Proof 2. Denote
. Is well-known that 
is an harmonical division, i.e. 
. Apply the Van Aubel's relation 
,
i.e.
.
PP8 (Ruben Dario Auqui). Let with the incircle 
and 
. Prove that 
.
Proof 1. Observe that
and 
.
Otherwise.
is h.d. So is the midpoint of 
.
In conclusion,
.
Proof 2.
.
Remark. I used the equation
and 
.
PP9. An acute with orthocenter 
, incenter 
, circumcircle 
and 
. Prove 
.
Proof. Denote the incircle
and the diameter ![$[TN]$](//latex.artofproblemsolving.com/7/7/d/77d65e7ebc3243ca10680055dee76914f5ffaf55.png)
of the circumcircle . Therefore:
is an isosceles trapezoid 
.
.
Otherwise. Denote
and 
. Thus, 
is cyclically 
a.s.o.
PP10 (Miguel Ochoa Sanchez). Let an-right , the midpoint of , the tangent points 
, 
and 
of its incircle and 
. Define 
and 
. Prove that 
.
Proof. Is well-known that in there is the relation 
and 
. Thus, 
, i.e.
, 
are -isosceles and
.
General case. Let. Is well-known that . Let 
. Thus, 
. Apply Theorem of Sines in 
. Since 
. Using 
obtain the relation 
.
Let
. Thus, 
, i.e. 
.
Obtain analogously that
, where 
. Thus, 
. In conclusion, 
. Prove easily that
. Observe that 
. Try to show 
.
PP11 (Francisco Garcia Capitan). Let with 
, where have the midpoint of , the tangent point 
of the-excircle with the line 
and 
so that 
. Prove that 
.
Proof. Is well-known that
and 
. Let 
. Thus, 
. So 
.
PP12. Let a
-right , where have the midpoint of , the tangent point of the -excircle with and 
so that . Prove that 
.
Proof. Is well-known that and . Let . Thus, 
. So 
- truly. If , 
are the midpoints of ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
, ![$[NT]$](//latex.artofproblemsolving.com/8/5/9/859d0d95f62b1064e76b2a99f19fa02921d5c192.png)
, then prove generally that 
doesn't separate 
and 
is parallel with the -bisector.
PP13. Let an isosceles trapezoid
with 
, its circumcircle 
and 
so that 
is cyclic. Prove that 
.
Proof.
. Otherwise. 
a.s.o.
PP14. Let with 
, 
and 
. Let 
be the circle with the center 
, 
and which is tangent to 
. Find 
.
Proof 1. Show easily that is acute. Denote the projection 
of on and . Since 
, i.e. 
obtain that
![$h_a(a-x)=2[ADC]=xb\iff$](//latex.artofproblemsolving.com/4/a/9/4a9b0cb4e2e33bebb0e1ada007bf62cc4c5dc0d8.png)
. Thus, denote 
and 
and
. Therefore, 
, i.e. 
. Obtain easily that 
. Otherwise, can
use the Heron's formula
. Indeed, 
and 
. Thus,
, i.e. 
. In conclusion, using the relation obtain that 
.
Proof 2. Observe that
. Denote: 
and 
, i.e. 
; 
and
, i.e. 
. Thus, 
. Apply the power of w.r.t. 
. Since 
obtain that 
. From the relations 
and eliminate 
and get 
. From get ![$7(7-2x)=\left[\frac {11(7-x)}{16}\right]^2\iff$](//latex.artofproblemsolving.com/a/4/d/a4db21a85c50ec9d3b5198e455e164b0290099f1.png)
.
But
. Hence 
.
Proof 3. Denote
, where 
, i.e. 
. Apply the Stewart's theorem to the cevian 
. Thus, 
. Thus, 
. Since obtain that
![$\left[\frac {11(7-x)}{16}\right]^2\iff$](//latex.artofproblemsolving.com/3/1/6/316d8b893e91daf92541093fb9ebc519ceddaa28.png)
.
But . Hence 
.
PP15 (Finala conc. Saraghin-2015). Let with the circumcentre 
. Denote the midpoints , 
and 
of the
sides , ![$[CA]$](//latex.artofproblemsolving.com/4/5/c/45c1acd47628de406680d04c09fe6314c3847acf.png)
and ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
respectively. Define the points 
. Prove that 
.
Proof.
.
PP16 (M. Ochoa Sanchez). Let an-isosceles with 
. Exists an interior point for which 
. Prove that 
.
Proof.
. Thus, 
. Apply the trigonometric form of the Ceva's theorem
. Denote 
and 
. Apply the generalized Pythagoras'
theorem to the side
. Apply the theorem of Sines in 
. Denote 
so that 
. Observe that 
.
In conclusion,
(Angel Lazo HK & V.N.).
PP17 (Ruben Dario). Let an equilateral with the circumcircle 
. For 
let 
. Prove that 
.
Proof 1. Suppose w.l.o.g. that
. Denote 
and 
. The Pompeiu's theorem 
, i.e. 
is well-known or can prove
easily it. In this case prove easily that
. Apply the well-known identity in any 
for the
triangles
and 
. Since 
obtain that the required relation .
Remark. Generalized Pythagoras' theorem
(Pompeiu's).
The relation
![$\left[\left(n^2+p^2\right)+(np)\right]^2\implies$](//latex.artofproblemsolving.com/c/4/3/c4304553095d777762114f020229d701904d749f.png)
.
Proof 2.![$[ASB]+[ASC]=[ABSC]=[BCA]+[BCS]\iff$](//latex.artofproblemsolving.com/d/b/c/dbcbc8d8be43f7a45c86d2dbab11d0771354db7e.png)
On other hand,
![$[DSE]+[DSF]=[ESF]\iff$](//latex.artofproblemsolving.com/a/f/5/af549e2865f1fc53a91115bafd9d81e1df554dd7.png)
From the
relation obtain that 
.
PP18 (Ruben Dario). Let an equilateral with the circumcircle . For a point 
denote
its projections , , on the tangent 
in the point to 
. Prove that 
.
Proof (Miguel Ochoa Sanchez). Suppose w.l.o.g.
. Denote 
and 
. Thus, 
and 
.
Thus
. In conclusion, 
![$4R^2\left[\sin^4\phi+\sin^4(60^{\circ}+\phi )+\sin^4(60^{\circ}-\phi )\right]$](//latex.artofproblemsolving.com/a/c/d/acdc03b7cad6591664c39fa4315ad10fadaca775.png)
.
Since
and 
obtain that the required identity .
Remark.
![$4E=(1-\cos 2\phi )^2+\left[1-\cos (120^{\circ}+2\phi )\right]^2+\left[1-\cos (120^{\circ}-2\phi )\right]^2=$](//latex.artofproblemsolving.com/d/9/9/d993e988de46811d9262a7d39695f87a73d5d0d9.png)
![$\left[1+\cos (60^{\circ}-2\phi )\right]^2+\left[1+\cos (60^{\circ}+2\phi )\right]^2=$](//latex.artofproblemsolving.com/3/7/8/3780f9b6f7fb13297c04dc0ece66283c20e26898.png)
![$3+\cos^22\phi +\sin^2(30^{\circ}+2\phi )+\sin^2(30^{\circ}-2\phi )+2\left[\cos (60-2\phi )+\cos (60+2\phi )-\cos 2\phi\right]\iff$](//latex.artofproblemsolving.com/b/e/5/be5568d6c0c03aec1db7987ce0433bf82cfd57f8.png)
![$8E=6+1+\cos 4\phi +\left[1-\cos (60^{\circ}+4\phi )\right]+\left[1-\cos (60^{\circ}-4\phi )\right]=$](//latex.artofproblemsolving.com/5/4/5/5450d1f7f5e753d361b16787e59e723d6cbdb245.png)
, i.e. the identity .
PP19 (Miguel Ochoa Sanchez). Let
be a trapezoid so that 
and 
. Denote
and suppose that there is 
so that 
. Prove that 
Proof.
PP20 (IMO ShortList 2003). Let an isosceles with 
, whose incentre is . Let be a point on the circumcircle of the triangle 
lying inside . The lines through parallel
to
and 
meet 
at and , respectively. The line through parallel to meets and at and 
, respectively. Prove that 
and 
intersect on the circumcircle of 
Commentary. The enunciation of this problem conceals vainly in its debut that the circumcircle of
is in fact the circle with the diameter 
where 
is the 
- exincenter of the triangle 
Proof. Let : the circumcircle of ; the circle 
with the diameter ![$[II_{c}]$](//latex.artofproblemsolving.com/4/d/3/4d360c2d7cf645c4c380fd8f59ddcee5827c26fd.png)
; the circumcircles 
, 
of the isosceles trapezoids 
, 
respectively ;
the second intersection point between the line 
and the circle ; the intersection 
Prove easily that the lines , are the tangents from
to the circle 
From the relation 
obtain 
i.e. the quadrilaterals 
, 
are similarly as 
Thus,
Variation on the same theme. Let
be a fixed isosceles triangle 
and let 
, 
be two mobile points so that 
and exists a point 
for which
Construct the points 
for which 
Ascertain the geometrical locus of the intersection 
-right 
with the -altitude 
, where 
and 
. Denote 
the incircle 
of
and the incircle 
of 
the tangent line 
to 
at 
and the tangent line 
to 
at 
, where
the incircle 
of 
and the incircle 
of 
Prove that 
Proof. Let
. Thus, 
and 
PP2 (Ruben Dario). Let
-right with 
and altitude , where . Prove that 
so that 
with 
and in thiscase, if denote
and 
, then 
and 
are the bisectors of 
and 
.Proof 1. Apply the Ceva's theorem
. Apply the theorem of cathetus 
The difference of relations
and 
, i.e. 
. Exists at least one 
so that 
In conclusion, there is only one positive root 
, i.e. there is only one 
which respects mentionedconditions. Thus,
and 
, i.e. the ray 
is the bisector of Proof 2. Apply the Ceva's theorem
. Apply the theorem of the symmedian 
Hence 
a.s.o.Remark.
the ray is the bisector of . Let 
so that
and 
. Prove easily that 
and 
. Hence 
and 
.An easy extension. Let
with -symmedian , where . Suppose that exists so that 
Let
. Prove that 
, 
and 
PP3 (M. Ochoa Sanchez). Let an acute
with orthocenter 
circumcircle 
and 
Prove that 
Proof. Suppose w.l.o.g.
and let 
. Prove easily 
. Apply the theorem of Sinesin
the relation 
.Remark.
is equilateral.PP4 (M. Ochoa Sanchez). Let
with 
and so that 
. Prove that 
. Proof. Denote
. Apply the theorem of Sines in the triangles 
.PP5. Let
with 
. Prove that 
Proof. Suppose w.l.o.g.
. Let 
and 
. Is well-known that 
and 
. Therefore 
.
.L1.
there is the equivalence 
(standard notations).Proof. Observe that
a.s.o. Thus, 
.Remark 1. Denote the midpoint
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the incircle 
and the -excircle 
. Is well-known that is the midpoint of
, the midpoint 
(south) of ![$[II_a]$](http://latex.artofproblemsolving.com/b/9/d/b9d69f173ce4bad72631095440791204d2b021e3.png)
belongs to the circumcircle of and from the trapezoid 
obtain that 
,where
is the length of the arrow of 
. Prove easily that the Ruben Dario's equivalence 
.Remark 2.
![$\frac 14\cdot\left[\frac {s^2}{(s-b)(s-c)}-\frac {as}{(s-b)(s-c)}+1\right]=$](http://latex.artofproblemsolving.com/a/e/9/ae9b6b3aacf77f776cd31605c2eda916ba83a253.png)
![$\frac 14\cdot\left[1+\frac {s(s-a)}{(s-b)(s-c)}\right]=$](http://latex.artofproblemsolving.com/3/5/3/353f474e7db769c22aa63e39cfa43237187171b6.png)
.PP6 (Ruben Dario). Let
with the incircle 
which touches at 
, 
and 
. Let 
. Prove that 
.Proof. Prove easily or is well-known that
and 
. Thus, the points 
belong to the circle with the dameter ![$[AI]$](http://latex.artofproblemsolving.com/c/0/8/c08892aba53b466068fea586787c79e1aa04654a.png)
.In conclusion,
. Nice problem!P7 (Ruben Dario). Let
with 
. Prove that 
.Proof 1. Apply Menelaus' theorem to
.Proof 2. Denote
. Is well-known that 
is an harmonical division, i.e. 
. Apply the Van Aubel's relation 
,i.e.
.PP8 (Ruben Dario Auqui). Let
with the incircle 
and 
. Prove that 
.Proof 1. Observe that
and 
.Otherwise.
is h.d. So is the midpoint of 
.In conclusion,
.Proof 2.
.Remark. I used the equation
and 
.PP9. An acute
with orthocenter 
, incenter 
, circumcircle 
and 
. Prove 
. Proof. Denote the incircle
and the diameter ![$[TN]$](http://latex.artofproblemsolving.com/7/7/d/77d65e7ebc3243ca10680055dee76914f5ffaf55.png)
of the circumcircle . Therefore:
is an isosceles trapezoid 
.
.Otherwise. Denote
and 
. Thus, 
is cyclically 
a.s.o. PP10 (Miguel Ochoa Sanchez). Let an
-right , the midpoint of , the tangent points 
, 
and 
of its incircle
and 
. Define 
and 
. Prove that 
. Proof. Is well-known that in
there is the relation 
and 
. Thus, 
, i.e.
, 
are -isosceles and
.General case. Let
. Is well-known that . Let 
. Thus, 
. Apply Theorem of Sines in 
. Since 
. Using 
obtain the relation 
.Let
. Thus, 
, i.e. 
.Obtain analogously that
, where 
. Thus, 
. In conclusion, 
. Prove easily that
. Observe that 
. Try to show 
.PP11 (Francisco Garcia Capitan). Let
with 
, where have the midpoint of , the tangent point 
of the
-excircle with the line 
and 
so that 
. Prove that 
.Proof. Is well-known that
and 
. Let 
. Thus, 
. So 
. PP12. Let a
-right , where have the midpoint of , the tangent point of the -excircle with and 
so that . Prove that 
. Proof. Is well-known that
and . Let . Thus, 
. So 
- truly. If , 
are the midpoints of ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
, ![$[NT]$](http://latex.artofproblemsolving.com/8/5/9/859d0d95f62b1064e76b2a99f19fa02921d5c192.png)
, then prove generally that 
doesn't separate 
and 
is parallel with the -bisector. PP13. Let an isosceles trapezoid
with 
, its circumcircle 
and 
so that 
is cyclic. Prove that 
. Proof.
. Otherwise. 
a.s.o.PP14. Let
with 
, 
and 
. Let 
be the circle with the center 
, 
and which is tangent to 
. Find 
.Proof 1. Show easily that
is acute. Denote the projection 
of on and . Since 
, i.e. 
obtain that![$h_a(a-x)=2[ADC]=xb\iff$](http://latex.artofproblemsolving.com/4/a/9/4a9b0cb4e2e33bebb0e1ada007bf62cc4c5dc0d8.png)
. Thus, denote 
and 
and
. Therefore, 
, i.e. 
. Obtain easily that 
. Otherwise, canuse the Heron's formula
. Indeed, 
and 
. Thus,
, i.e. 
. In conclusion, using the relation obtain that 
.Proof 2. Observe that
. Denote: 
and 
, i.e. 
; 
and
, i.e. 
. Thus, 
. Apply the power of w.r.t. 
. Since 
obtain that 
. From the relations 
and
eliminate 
and get 
. From get ![$7(7-2x)=\left[\frac {11(7-x)}{16}\right]^2\iff$](http://latex.artofproblemsolving.com/a/4/d/a4db21a85c50ec9d3b5198e455e164b0290099f1.png)
.But
. Hence 
.Proof 3. Denote
, where 
, i.e. 
. Apply the Stewart's theorem to the cevian 
. Thus, 
. Thus, 
. Since obtain that 
![$\left[\frac {11(7-x)}{16}\right]^2\iff$](http://latex.artofproblemsolving.com/3/1/6/316d8b893e91daf92541093fb9ebc519ceddaa28.png)
.But
. Hence 
. PP15 (Finala conc. Saraghin-2015). Let
with the circumcentre 
. Denote the midpoints , 
and 
of thesides
, ![$[CA]$](http://latex.artofproblemsolving.com/4/5/c/45c1acd47628de406680d04c09fe6314c3847acf.png)
and ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
respectively. Define the points 
. Prove that 
. Proof.
.PP16 (M. Ochoa Sanchez). Let an
-isosceles with 
. Exists an interior point for which 
. Prove that 
. Proof.
. Thus, 
. Apply the trigonometric form of the Ceva's theorem 
. Denote 
and 
. Apply the generalized Pythagoras'theorem to the side
. Apply the theorem of Sines in 
. Denote 
so that 
. Observe that 
.In conclusion,
(Angel Lazo HK & V.N.). PP17 (Ruben Dario). Let an equilateral
with the circumcircle 
. For 
let 
. Prove that 
. Proof 1. Suppose w.l.o.g. that
. Denote 
and 
. The Pompeiu's theorem 
, i.e. 
is well-known or can proveeasily it. In this case prove easily that
. Apply the well-known identity in any 
for thetriangles
and 
. Since 
obtain that the required relation .Remark. Generalized Pythagoras' theorem
(Pompeiu's).The relation
![$\left[\left(n^2+p^2\right)+(np)\right]^2\implies$](http://latex.artofproblemsolving.com/c/4/3/c4304553095d777762114f020229d701904d749f.png)
.Proof 2.
![$[ASB]+[ASC]=[ABSC]=[BCA]+[BCS]\iff$](http://latex.artofproblemsolving.com/d/b/c/dbcbc8d8be43f7a45c86d2dbab11d0771354db7e.png)
On other hand,![$[DSE]+[DSF]=[ESF]\iff$](http://latex.artofproblemsolving.com/a/f/5/af549e2865f1fc53a91115bafd9d81e1df554dd7.png)
From therelation
obtain that 
.PP18 (Ruben Dario). Let an equilateral
with the circumcircle . For a point 
denoteits projections
, , on the tangent 
in the point to 
. Prove that 
. Proof (Miguel Ochoa Sanchez). Suppose w.l.o.g.
. Denote 
and 
. Thus, 
and 
.Thus
. In conclusion, 
![$4R^2\left[\sin^4\phi+\sin^4(60^{\circ}+\phi )+\sin^4(60^{\circ}-\phi )\right]$](http://latex.artofproblemsolving.com/a/c/d/acdc03b7cad6591664c39fa4315ad10fadaca775.png)
.Since
and 
obtain that the required identity .Remark.
![$4E=(1-\cos 2\phi )^2+\left[1-\cos (120^{\circ}+2\phi )\right]^2+\left[1-\cos (120^{\circ}-2\phi )\right]^2=$](http://latex.artofproblemsolving.com/d/9/9/d993e988de46811d9262a7d39695f87a73d5d0d9.png)
![$\left[1+\cos (60^{\circ}-2\phi )\right]^2+\left[1+\cos (60^{\circ}+2\phi )\right]^2=$](http://latex.artofproblemsolving.com/3/7/8/3780f9b6f7fb13297c04dc0ece66283c20e26898.png)
![$3+\cos^22\phi +\sin^2(30^{\circ}+2\phi )+\sin^2(30^{\circ}-2\phi )+2\left[\cos (60-2\phi )+\cos (60+2\phi )-\cos 2\phi\right]\iff$](http://latex.artofproblemsolving.com/b/e/5/be5568d6c0c03aec1db7987ce0433bf82cfd57f8.png)
![$8E=6+1+\cos 4\phi +\left[1-\cos (60^{\circ}+4\phi )\right]+\left[1-\cos (60^{\circ}-4\phi )\right]=$](http://latex.artofproblemsolving.com/5/4/5/5450d1f7f5e753d361b16787e59e723d6cbdb245.png)
, i.e. the identity .PP19 (Miguel Ochoa Sanchez). Let
be a trapezoid so that 
and 
. Denote
and suppose that there is 
so that 
. Prove that 
Proof.
PP20 (IMO ShortList 2003). Let an isosceles
with 
, whose incentre is . Let be a point on the circumcircle of the triangle 
lying inside . The lines through parallelto
and 
meet 
at and , respectively. The line through parallel to meets and at and 
, respectively. Prove that 
and 
intersect on the circumcircle of 
Commentary. The enunciation of this problem conceals vainly in its debut that the circumcircle of
is in fact the circle with the diameter 
where 
is the 
- exincenter of the triangle 
Proof. Let : the circumcircle
of ; the circle 
with the diameter ![$[II_{c}]$](http://latex.artofproblemsolving.com/4/d/3/4d360c2d7cf645c4c380fd8f59ddcee5827c26fd.png)
; the circumcircles 
, 
of the isosceles trapezoids 
, 
respectively ;the second intersection point
between the line 
and the circle ; the intersection 
Prove easily that the lines , are the tangents from to the circle 
From the relation 
obtain 
i.e. the quadrilaterals 
, 
are similarly as 
Thus,
Variation on the same theme. Let
be a fixed isosceles triangle 
and let 
, 
be two mobile points so that 
and exists a point 
for which
Construct the points 
for which 
Ascertain the geometrical locus of the intersection 
where the point 
is the 
This post has been edited 274 times. Last edited by Virgil Nicula, Mar 13, 2018, 7:44 AM
August 2018
January 2018