400. Some nice geometry problems for the high school I.
by Virgil Nicula, Oct 29, 2014, 11:14 AM
PP8. Let 
and for 
let its projections 
on 
respectively, 
. Circumcircle of 
cut again 
in 
. Prove that 
Particular cases..
- the orthocenter of . In this case is the midpoint of the side ![$[BC]\implies RH\perp AM$](//latex.artofproblemsolving.com/8/4/8/848d951b0866e88029aaaa4359c652d1d318e56b.png)
.
- the incenter of . In this case is the point 
.
Proof 1. Denote the circumcircles
of 
respectively. Denote 
. Observe that 
and
. In conclusion, the point 
is the radical center of the circles 
. Since ![$[AP]$](//latex.artofproblemsolving.com/f/6/2/f6235de9bde9e47652ee5f3cfbad1a2666873220.png)
is the diameter of 
obtain that 
.
Proof 2. Denote the arrow
of the arc 
for which 
a.s.o. Define analogously 
and 
. I"ll use the well-known identities 
.
PP9. Let with incircle 
. Let 
, the Gergonne point 
and the Nagel point 
. Prove that 
.
Proof 1. Suppose w.l.o.g. that
and denote : the orthocenter 
and 
for which 
.
Show easily that
and 
. Therefore, 
![$\left[(b+c)-a\right]^2=\left[a+(b+c)\right]\left[3a-(b+c)\right]\iff$](//latex.artofproblemsolving.com/4/2/5/425d6bca65820ba3f77ceb12e93c98f16b7c3d84.png)
.
Proof 2. Suppose w.l.o.g. that . Is well-known that 
. Show easily that 
. Thus 
the division
is harmonically 
the division 
is harmonically 
.
PP10. Let an acute with incircle 
and circumcircle 
. Let 
for which 
. Prove that 
Proof.
. Thus, 
. In conclusion, 
.
Extension. Let an acute with incircle 
and circumcircle . Let so that 
, 
and 
, 
.
Prove that
. Particular cases. 
what are evidently. Thus we verified the upper identity ! Also we can find easily the points
for which 
, i.e. 
.
Proof. Denote
for which 
. Apply the generalized Pytagoras' relation in 
: 
![$ \left[sx - (s - a)h_a\right]\left(x - h_a\right) + h_a(s - a)\left(2R - h_a\right) =$](//latex.artofproblemsolving.com/7/9/e/79eda64de53c040d2adf3e07dc1a231e7f8e83ea.png)
. Since 
obtain 
.
PP11. Let an
-right . Let the midpoint 
of ![$ [AC]$](//latex.artofproblemsolving.com/5/b/0/5b08b2b92472414250205866b93405af4772e86c.png)
and for which . Let 
and 
. Let 
be
the second intersection between
and the circle with the diameter . Prove that 
is cyclic and 
is incenter of the triangle 
.
Another similar problem. Let an-right . Denote the midpoint of ![$ [AB]$](//latex.artofproblemsolving.com/d/7/a/d7a8027c238eec9cf67de0f7ec6cb1df4df49a61.png)
and for which . The incircle and the 
-exincircle touch the sideline 
in 
, 
respectively. Let be the second intersection between and the circle with the diameter . Prove that the quadrilateral is cyclically and 
is incenter of the triangle .
Remark. Observe that
is a harmonical division because 
. The proof is simply, is a particular application of the harmonical division. For example, 
a.s.o. It is analogously and at the previous problem !
PP12. Let an acute with incircle 
and circumcircle 
. Circles 
and 
are tangent internally to the circle 
in the
same . The circle 
is tangent externally to 
and is tangent internally to 
. Prove that 
, where 
and ![$ S\equiv [ABC]$](//latex.artofproblemsolving.com/8/d/f/8df17304ca8c5a866f0e34f9089fe302737d3f65.png)
- the area of 
.
Proof. Prove easily that
. The relations 
are well-known.
Proof 1. Denote
, 
. Apply the Stewart's theorem in 
for the cevian-rays 
and 
.
Proof 2. Apply the Pythagoras' theorem
.
Therefore,
, i.e. 
. Observe that 
, where 
is the - exinradius of .
Remark. Prove easily that two more interesting relations
and 
, 
.
Observe that
![$\left\{\begin{array}{c}\boxed{\ r\le \sqrt [3]{\prod PQ}\le \frac{R}{2}\ }\\\\ \boxed{\ r\le\frac{1}{3}\cdot \sum PQ\le \frac{R^{2}}{4r}\ }\end{array}\right\|$](//latex.artofproblemsolving.com/c/f/d/cfd778ce432ccf113a47bac4192da7f28a466f37.png)
PP13. Let parallelogram
, 
, the symmetrical 
of 
w.r.t. 
and parallelogram 
with 
. Prove that 
and 
Proof. Let the origin
of the vectorial system, i.e. 
and 
Hence 
, 
, exists 
so that 
and
i.e. 
Thus, 
Hence
Since 
obtain that 
Thus,
.
PP14. Let be an interior point of for which 
. Consider the triangle 
for which 
. Prove that 
.
Proof. Let be the point so that 
separates and and 
. Thus, 
and 
. Apply the theorem of Cosines in 
a.s.o.
Particular case.
. Since 
apply the relation 
. In conclusion, 
.
PP15. Let an acute with circumcircle 
and the arrows 
. Find
the area![$S=[ABC]$](//latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
. Remark that 
.
Proof 1. I"ll use
a.s.o. Thus, ![$\left\{\begin{array}{cccccc}
\sum \left(r_a-r\right)=2\sum s_a=16 & \implies & (4R+r)-3r=16 & \implies & \boxed{2R=r+8} & (1)\\\\
\prod \left[s\left(r_a-r\right)\right]=abcr_ar_br_c & \implies & 20s^2=RS^2 & \implies & \boxed{Rr^2=20} & (2)\end{array}\right\|$](//latex.artofproblemsolving.com/b/0/9/b097ad83fb2cca7b231151e6c9610c1bed387a98.png)
.
From
and 
obtain easily that 
and 
. therefore, 
, 
, 
and 
the area of is 
.
Remark. Using Carnot's relation get
![$\sum\left[\delta_{BC}(O)+s_a\right]=\sum\delta_{BC}(O)+\sum s_a\iff$](//latex.artofproblemsolving.com/5/7/5/575a391d1f03ee3a1e82b069f282106ef1e20385.png)
. Thus, 
, where 
, i.e. .
PP16. Let an-right and 
so that 
. Let the midpoint of ![$[BE]$](//latex.artofproblemsolving.com/f/b/0/fb061a8a7c5f9403b5f9261840de9dfea7cb68cf.png)
and 
. Prove that 
.
Proof. Denote
. Observe that 
and 
. Apply the Menelaus" theorem
to the transversal
. Apply the generalized Pythagoras' theorem in the triangle 
. Thus, 
what is the thrue relation .
PP17. Let
be a circle with the diameter ![$[IJ]$](//latex.artofproblemsolving.com/f/f/d/ffd3dfe70945702b80b2e0dffb7c720a0880bbd3.png)
. For a point 
and a line 
so that 
denote 
the midpoints , 
of ![$[HI]$](//latex.artofproblemsolving.com/0/d/c/0dc940f1003c53129f8e322711c22f43fdabcd56.png)
, ![$[HJ]$](//latex.artofproblemsolving.com/3/1/f/31f889c6b2f674a1686719f840fb7a6a0538de98.png)
respectively
where
, 
the circles 
, 
so that 
and 
. Prove that 
.
Proof. Let the projection
of 
on . Since is the midpoint of and 
obtain that is the common midpoint of ![$[EF]$](//latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
and ![$[DG]$](//latex.artofproblemsolving.com/3/0/b/30bbd80f5722b9cd0703953eae2528558886b4df.png)
, i.e. .
PP18. Let with circumcircle 
. Prove that 
so that separates 
we have 
.
Proof 1. I"ll use the relation
for any (standard notations). Denote 
and 
. Thus, ![$[MAB]+[MAC]=[BAC]+[BMC]\iff$](//latex.artofproblemsolving.com/2/b/3/2b37a82f0597e3775fb2bfefb1575506bd651e56.png)
. Apply Ptolemy's theorem 
. In conclusion, 
, i.e. the ray 
is the -bisector of 
or is -isosceles.
Proof 2. Denote
, 
and 
. Thus, ![$[BAC]+[BMC]=[ABMC]\iff$](//latex.artofproblemsolving.com/6/0/b/60bb935fae5c63af571d8e29f126283b0a4017cf.png)
. In conclusion, 
.
and for 
let its projections 
on 
respectively, 
. Circumcircle of 
cut again 
in 
. Prove that 
Particular cases..
- the orthocenter of . In this case is the midpoint of the side ![$[BC]\implies RH\perp AM$](http://latex.artofproblemsolving.com/8/4/8/848d951b0866e88029aaaa4359c652d1d318e56b.png)
.
- the incenter of . In this case is the point 
.Proof 1. Denote the circumcircles
of 
respectively. Denote 
. Observe that 
and
. In conclusion, the point 
is the radical center of the circles 
. Since ![$[AP]$](http://latex.artofproblemsolving.com/f/6/2/f6235de9bde9e47652ee5f3cfbad1a2666873220.png)
is the diameter of 
obtain that 
.Proof 2. Denote the arrow
of the arc 
for which 
a.s.o. Define analogously 
and 
. I"ll use the well-known identities 
. PP9. Let
with incircle 
. Let 
, the Gergonne point 
and the Nagel point 
. Prove that 
.Proof 1. Suppose w.l.o.g. that
and denote : the orthocenter 
and 
for which 
.Show easily that
and 
. Therefore, 
![$\left[(b+c)-a\right]^2=\left[a+(b+c)\right]\left[3a-(b+c)\right]\iff$](http://latex.artofproblemsolving.com/4/2/5/425d6bca65820ba3f77ceb12e93c98f16b7c3d84.png)
.Proof 2. Suppose w.l.o.g. that
. Is well-known that 
. Show easily that 
. Thus 
the division
is harmonically 
the division 
is harmonically 
.PP10. Let an acute
with incircle 
and circumcircle 
. Let 
for which 
. Prove that 
Proof.
. Thus, 
. In conclusion, 
.Extension. Let an acute
with incircle 
and circumcircle . Let so that 
, 
and 
, 
.Prove that
. Particular cases. 
what are evidently. Thus we verified the upper identity ! Also we can find easily the points
for which 
, i.e. 
.Proof. Denote
for which 
. Apply the generalized Pytagoras' relation in 
: 
![$ \left[sx - (s - a)h_a\right]\left(x - h_a\right) + h_a(s - a)\left(2R - h_a\right) =$](http://latex.artofproblemsolving.com/7/9/e/79eda64de53c040d2adf3e07dc1a231e7f8e83ea.png)
. Since 
obtain 
.PP11. Let an
-right . Let the midpoint 
of ![$ [AC]$](http://latex.artofproblemsolving.com/5/b/0/5b08b2b92472414250205866b93405af4772e86c.png)
and for which . Let 
and 
. Let 
bethe second intersection between
and the circle with the diameter . Prove that 
is cyclic and 
is incenter of the triangle 
.Another similar problem. Let an
-right . Denote the midpoint of ![$ [AB]$](http://latex.artofproblemsolving.com/d/7/a/d7a8027c238eec9cf67de0f7ec6cb1df4df49a61.png)
and for which . The incircle and the 
-exincircle touch the sideline 
in 
, 
respectively. Let
be the second intersection between and the circle with the diameter . Prove that the quadrilateral is cyclically and 
is incenter of the triangle .Remark. Observe that
is a harmonical division because 
. The proof is simply, is a particular application of the harmonical division. For example, 
a.s.o. It is analogously and at the previous problem !PP12. Let an acute
with incircle 
and circumcircle 
. Circles 
and 
are tangent internally to the circle 
in thesame
. The circle 
is tangent externally to 
and is tangent internally to 
. Prove that 
, where 
and ![$ S\equiv [ABC]$](http://latex.artofproblemsolving.com/8/d/f/8df17304ca8c5a866f0e34f9089fe302737d3f65.png)
- the area of 
.Proof. Prove easily that
. The relations 
are well-known.Proof 1. Denote
, 
. Apply the Stewart's theorem in 
for the cevian-rays 
and 
.Proof 2. Apply the Pythagoras' theorem
.Therefore,
, i.e. 
. Observe that 
, where 
is the - exinradius of .Remark. Prove easily that two more interesting relations
and 
, 
.Observe that
![$\left\{\begin{array}{c}\boxed{\ r\le \sqrt [3]{\prod PQ}\le \frac{R}{2}\ }\\\\ \boxed{\ r\le\frac{1}{3}\cdot \sum PQ\le \frac{R^{2}}{4r}\ }\end{array}\right\|$](http://latex.artofproblemsolving.com/c/f/d/cfd778ce432ccf113a47bac4192da7f28a466f37.png)
PP13. Let parallelogram
, 
, the symmetrical 
of 
w.r.t. 
and parallelogram 
with 
. Prove that 
and 
Proof. Let the origin
of the vectorial system, i.e. 
and 
Hence 
, 
, exists 
so that 
and
i.e. 
Thus, 
Hence
Since 
obtain that 
Thus,
. PP14. Let
be an interior point of for which 
. Consider the triangle 
for which 
. Prove that 
.Proof. Let
be the point so that 
separates and and 
. Thus, 
and 
. Apply the theorem of Cosines in 
a.s.o.Particular case.
. Since 
apply the relation 
. In conclusion, 
.PP15. Let an acute
with circumcircle 
and the arrows 
. Findthe area
![$S=[ABC]$](http://latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
. Remark that 
.Proof 1. I"ll use
a.s.o. Thus, ![$\left\{\begin{array}{cccccc}
\sum \left(r_a-r\right)=2\sum s_a=16 & \implies & (4R+r)-3r=16 & \implies & \boxed{2R=r+8} & (1)\\\\
\prod \left[s\left(r_a-r\right)\right]=abcr_ar_br_c & \implies & 20s^2=RS^2 & \implies & \boxed{Rr^2=20} & (2)\end{array}\right\|$](http://latex.artofproblemsolving.com/b/0/9/b097ad83fb2cca7b231151e6c9610c1bed387a98.png)
.From
and 
obtain easily that 
and 
. therefore, 
, 
, 
and 
the area of is 
.Remark. Using Carnot's relation get
![$\sum\left[\delta_{BC}(O)+s_a\right]=\sum\delta_{BC}(O)+\sum s_a\iff$](http://latex.artofproblemsolving.com/5/7/5/575a391d1f03ee3a1e82b069f282106ef1e20385.png)
. Thus, 
, where 
, i.e. . PP16. Let an
-right and 
so that 
. Let the midpoint of ![$[BE]$](http://latex.artofproblemsolving.com/f/b/0/fb061a8a7c5f9403b5f9261840de9dfea7cb68cf.png)
and 
. Prove that 
.Proof. Denote
. Observe that 
and 
. Apply the Menelaus" theoremto the transversal
. Apply the generalized Pythagoras' theorem in the triangle 
. Thus, 
what is the thrue relation .PP17. Let
be a circle with the diameter ![$[IJ]$](http://latex.artofproblemsolving.com/f/f/d/ffd3dfe70945702b80b2e0dffb7c720a0880bbd3.png)
. For a point 
and a line 
so that 
denote 
the midpoints , 
of ![$[HI]$](http://latex.artofproblemsolving.com/0/d/c/0dc940f1003c53129f8e322711c22f43fdabcd56.png)
, ![$[HJ]$](http://latex.artofproblemsolving.com/3/1/f/31f889c6b2f674a1686719f840fb7a6a0538de98.png)
respectivelywhere
, 
the circles 
, 
so that 
and 
. Prove that 
.Proof. Let the projection
of 
on . Since is the midpoint of and 
obtain that is the common midpoint of ![$[EF]$](http://latex.artofproblemsolving.com/7/6/3/763239c0ce4fccc63411d3d6cb0011f7f6cc3a31.png)
and ![$[DG]$](http://latex.artofproblemsolving.com/3/0/b/30bbd80f5722b9cd0703953eae2528558886b4df.png)
, i.e. . PP18. Let
with circumcircle 
. Prove that 
so that separates 
we have 
. Proof 1. I"ll use the relation
for any (standard notations). Denote 
and 
. Thus, ![$[MAB]+[MAC]=[BAC]+[BMC]\iff$](http://latex.artofproblemsolving.com/2/b/3/2b37a82f0597e3775fb2bfefb1575506bd651e56.png)
. Apply Ptolemy's theorem 
. In conclusion, 
, i.e. the ray 
is the -bisector of 
or is -isosceles.Proof 2. Denote
, 
and 
. Thus, ![$[BAC]+[BMC]=[ABMC]\iff$](http://latex.artofproblemsolving.com/6/0/b/60bb935fae5c63af571d8e29f126283b0a4017cf.png)
. In conclusion, 
.This post has been edited 222 times. Last edited by Virgil Nicula, Feb 23, 2016, 6:44 AM
August 2018
January 2018