156. Some problems with perpendicularities.
by Virgil Nicula, Oct 15, 2010, 1:42 PM
Quote:
Proposed problem 1. Let 
be a triangle with the 
-angle bisector 
and the -median 
, where 
. Consider 
so that
and 
so that 
. Prove that the point 
belongs to the circumcircle of the triangle , i.e. 
.
be a triangle with the 
-angle bisector 
and the -median 
, where 
. Consider 
so that
and 
so that 
. Prove that the point 
belongs to the circumcircle of the triangle , i.e. 
.
of 
with the circumcircle of the given triangle. From the well-known properties 
,
and 
obtain the chain of equalities 
, i.e. 
.Proof 2 (metric). Denote
and 
for which 
, i.e. the point is the second intersection of with the circumcircle of 
.Denote
. Apply Menelaus' theorem to 
.Observe that
from where obtain 
. From the product of the relations 
obtain that
. In conclusion, 
, i.e. . Quote:
Proposed problem 2. Let be a triangle. Let 
and 
be the points in which the -median and the -angle bisector meet the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
respectively. Let
and 
for which 
and 
for which 
. Prove that 
.
be a triangle. Let 
and 
be the points in which the -median and the -angle bisector meet the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
respectively. Let
and 
for which 
and 
for which 
. Prove that 
. Proof. Denote the second intersection
of 
with the circumcircle of and 
. The Simson line 
of w.r.t.
is perpendicular to 
. From 
and 
obtain that 
. Hence 
.Quote:
Proposed problem 3. Fie trapezul 
with 
and 
, Consider a line 
so that
. The circle 
cut 
in 
. Prove that the point is the orthocenter of 
.
with 
and 
, Consider a line 
so that
. The circle 
cut 
in 
. Prove that the point is the orthocenter of 
.
. Prove easily that 
, 
, 
and 
. 
is a parallelogramobtain that
In conclusion, 
and 
, i.e. the point is the orthocenter of .Proposed problem 4. Let
be a triangle with the circumcircle 
and the incircle 
.Define the points
. Prove that 
and 
.Proof. Denote
- the power of the point 
w.r.t. the circle 
. Thus, 
.The relations
a.s.o. are well-known. Apply the Stewart's theorem to the rays 
, 
in 
, 
respectively :
. In conclusion, 
, i.e. .Let
and 
be the orthogonal projections of 
on the perpendicular bisectors of and ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
respectively; let 
and 
be the midpoints of and respectively;let
and 
be the orthogonal projections of on 
and 
respectively, i. e. the points where the incircle of triangle touches the sides and . It is well-knownthat
, where 
is the semiperimeter of . Thus, 
. Since the quadrilateral 
is a rectangle (it has three right angles: at
, at and at ), we have 
, so 
. Now we have 
; therefore,
. Similarly, 
. Finally, we have 
and 
, i. e. 
, so that 
. Similarly, 
. Hence,
, since the sides of the two angles are respectively parallel. Together with 
and , this shows that 
are similar,with the ratio of similitude
. Hence, 
. Now, since 
and 
, the points and lie on the circle with diameter 
.In other words, the circumcircle of triangle
has ![$[OI]$](http://latex.artofproblemsolving.com/1/e/6/1e6d88a33008d2bd15b8d0f7de4ff31d77c52d0f.png)
as diameter. Therefore, after the extended law of sines, 
. Hence we have
, and thus 
. PP5. In
Denote 
and 
such that 
. Prove that 
.Proof 1. Denote
and 
so that 
. Observe that 
. Thus, 
and the points 
, are homologously w.r.t. this similarity. Therefore,
, i.e. the quadrilateral 
is cyclically. In conclusion, 
, i.e. 
.Proof 2.
PP6 (old problem). Let
be a rectangle and let 
be the projection of 
on the diagonal 
. Consider two points 
and 
so that 
. Prove that 
.Proof 1. Define the point
so that 
and 
. Observe that 
and 
. Therefore, the points and are homologously in the similarity 
,i.e.
the quadrilateral 
is inscribed in the circle with the diameter ![$[BC]\implies CZ\perp BX$](http://latex.artofproblemsolving.com/c/1/0/c10f35bb520a2de78aa242b8de46af58d0e4d836.png)
. Since 
and
obtain that 
. Since 
obtain that 
. In conclusion, 
is a parallelogram 
. Since 
obtain that .Proof 2 (analytic). Let
. Thus, 
. Denote 
. Therefore,
. Thus, 
![$\frac {b\cdot\left[ \frac {(1-\lambda )b^2}{a^2+b^2}-1\right]}{ab^2\cdot \frac {1-\lambda}{a^2+b^2}}$](http://latex.artofproblemsolving.com/1/e/1/1e1d6513bdbd8774c3f43f13ccf6a95d952d4b9e.png)
![$\cdot\frac {(1-\lambda )b\cdot \left[\frac {b^2}{a^2+b^2}-1\right]}{a\cdot \left[\frac {(1-\lambda )b^2}{a^2+b^2}-1\right]}=-1$](http://latex.artofproblemsolving.com/3/c/c/3ccbc64794166bda4ce0b98c904edc2f81eb223b.png)
, what is true.PP7. Let
with the incircle 
. Denote the midpoint of . Prove that 
Proof 1 (metric).
.Proof 2 (metric). Denote
and suppose that 
. In this case 
. Prove analogously that in the second case exists the equivalence 
.Proof 3 (synthetic). Denote
and suppose that . Observe that 
.Proof 4 (synthetic). Suppose
. Denote the midpoint of ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and 
. Observe that 
.Proof 5 (Yetty -synthetic). Suppose
and denote , the diameter ![$[TN]$](http://latex.artofproblemsolving.com/7/7/d/77d65e7ebc3243ca10680055dee76914f5ffaf55.png)
of 
and 
. Is well-known that 
belongs ro the -exincircleand
. In conclusion, 
is -isosceles 
.Proof 6 (underzero -synthetic). Denote
, 
and 
. Is well-known that 
. Thus, 
, i.e, is the midpoint of ![$[BP]$](http://latex.artofproblemsolving.com/c/d/a/cda0272ae887e314ff58215fbf55333fa558fefd.png)
. Therefore, 
.PP8 (Ochoa Sanchez). Let
with the incircle 
and 
so that 
. Proof.
.This post has been edited 84 times. Last edited by Virgil Nicula, Nov 22, 2015, 9:11 PM
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