384. Geometry problems (I).
by Virgil Nicula, Aug 12, 2013, 3:18 PM
PP11 (Ruben Dario). Let 
-right 
with incircle 
which touches it at 
. Let 
. Prove that 
.
Proof 1. Let
and observe that 
. Thus, 
. Denote 
and observe that 
.
Proof 2. 
.Since 
, obtain that 
![$\boxed{c\left[c^2+(s-a)^2\right]=s(s-b)^2}\ (*)$](//latex.artofproblemsolving.com/4/5/2/4529f099a82892100ad8635bb38055747f0f19eb.png)
.
Denote
. Thus, 
and 
. The relation 
becomes 
. Thus, 
. In conclusion, 
.
An easy extension. Let be a triangle with 
. The its incircle touches it at 
, 
and
. Denote . Prove that 
.
Proof. Denote and 
. Thus, 
. Therefore,
.
PP12 (Elberling Vargas Diaz). Let
be a square with center 
. Denote the midpoints 
, 
, 
of 
, 
, 
respectively. Denote
and 
so that the line 
is tangent to the incircle 
of the given square. Denote 
. Find 
.
Proof. Suppose w.l.o.g. that
. Denote 
and 
. Thus, is tangent to 
, i.e.
, i.e. 
. Thus, 
.
An easy extension. Let be a square with center . Denote the midpoints , , and 
of , , and 
respectively. Denote ,
, 
, 
so that and 
are tangent to the incircle of the given square. Denote 
. Prove that 
.
Proof. Suppose w.l.o.g. that . Denote 
. Therefore,
. Observe that
. In conclusion, 
.
PP13. Let two lines
and 
(concurrent or parallel) what are tangent to a circle 
in its points and 
respectively. Prove that
the chain 
.
Proof.
.
Particular cases.
Let with incircle what touches 
at 
and 
. Prove that 
.
Let with incircle what touches 
at 
. Denote 
. Prove that 
and 
.
PP14 (Silver Samuel Palacios). Let so that 
and 
. Consider 
for which construct the rectangle 
with 
and
. Denote 
, 
so that 
, 
and 
. Prove that 
and 
.
Proof. Denote
and 
. Thus, 
. Apply the Menelaus' theorem to:
.
.
.
. Therefore,
and 
. Thus, 
.
Apply the Menelaus' theorem to
.
PP15. Let convex with 
. Denote 
. Prove that 
.
Proof. Prove easily that
. In conclusion, . Nice and easy problem !
PP16. Let an equilateral with circumcircle . For 
so that separates and 
denote the 
. Prove that ![$[QPR]=2\cdot [ABC]$](//latex.artofproblemsolving.com/2/8/2/2822583080d0a905e17868367133dd1b124946e0.png)
.
Proof. Suppose w.l.o.g. that
and denote 
where 
. Thus, 
and![$[ABC]=\frac {\sqrt 3}{4}$](//latex.artofproblemsolving.com/3/7/2/372282ac1d07c3624ff15b7dda80514c707813db.png)
. Prove easily that 
and 
. Apply Menelaus' theorem to the transvesals: 
.
Thus,![$[QPR]=[QAR]-[QAP]-[RAP]\iff$](//latex.artofproblemsolving.com/f/6/0/f605c0da15fa1599d8454d3b982cfbe7f63bbf42.png)
![$2\cdot [QPR]=AQ\cdot AR\cdot\sin\widehat{BAC}-AQ\cdot BP\cdot\sin\widehat{ABC}-AR\cdot CP\cdot\sin\widehat{ACB}=$](//latex.artofproblemsolving.com/f/e/c/fec97ed25fdf43860289695e83ab42588c84f770.png)
![$4\cdot [ABC]\implies$](//latex.artofproblemsolving.com/e/d/c/edc6e049b4399a29916886408f105673d8df012b.png)
.
An easy extension. Let with the circumcircle . For so that separates and denote
so that doesn't separate 
and 
. Let 
and 
. Prove that ![$[QPR]=\frac {2axy\left(b^2x+c^2y\right)}{\left[b^2x+\left(c^2-a^2\right)y\right]\left[c^2y+\left(b^2-a^2\right)x\right]}\cdot [ABC]$](//latex.artofproblemsolving.com/6/f/f/6ffb4d6708f6fc947d6093749f2d914c9ea174e8.png)
.
Proof. Apply the Stewart's relation to
in 
. Thus, 
and 
. Apply the Menelaus' theorem:
. Therefore, ![$2\cdot [QPR]=AQ\cdot AR\cdot\sin\widehat{BAC}-$](//latex.artofproblemsolving.com/3/3/4/3340d14adb222416c6506371c6ffdcdf9dff955e.png)
![$\frac{bc\left(b^2x+c^2y\right)^2\sin A}{\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]} -$](//latex.artofproblemsolving.com/a/c/4/ac4b79510e1109b11b6178cc60428bf4fafd1500.png)
![$\frac {bcx\left(b^2x+c^2y\right)}{2R\left[c^2y+\left(b^2-a^2\right)x\right]}-$](//latex.artofproblemsolving.com/d/8/4/d840d4d5b41a550970bef00a3a3badf21cfb4e60.png)
![$\frac {bcy\left(b^2x+c^2y\right)}{2R\left[b^2x+\left(c^2-a^2\right)y\right]}=$](//latex.artofproblemsolving.com/a/5/0/a50fa687882d70e5201d5a82366308c89f0669f6.png)
![$\frac{abc\left(b^2x+c^2y\right)^2-bcx\left(b^2x+c^2y\right)\left[b^2x+\left(c^2-a^2\right)y\right]-bcy\left(b^2x+c^2y\right)\left[c^2y+\left(b^2-a^2\right)x\right]}{2R\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]}=$](//latex.artofproblemsolving.com/3/c/7/3c7de710d7df08c942bdf13e4350b5078b56c7c2.png)
![$\frac {bc\left(b^2x+c^2y\right)}{2R\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]}\cdot$](//latex.artofproblemsolving.com/0/6/7/067d6939e960873d559d16f78be7c5a160bb7a38.png)
![$\left\{a\left(b^2x+c^2y\right)-x\left[b^2x+\left(c^2-a^2\right)y\right]-y\left[c^2y+\left(b^2-a^2\right)x\right]\right\}=$](//latex.artofproblemsolving.com/7/7/0/7705ff4eda9afa1c215f51ad044c4d5e935c77b5.png)
![$\frac {h_a\left(b^2x+c^2y\right)}{\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]}$](//latex.artofproblemsolving.com/d/c/c/dcc84317853e28b9ee0cac305600b2d138d87bb8.png)
![$\cdot\left[b^2x(a-x)+c^2y(a-y)+xy\left(2a^2-b^2-c^2\right)\right]=$](//latex.artofproblemsolving.com/f/d/3/fd3f433fd70ed7f6bbeb7d83846fb9ea308b4ac6.png)
![$\frac {2a^2h_axy\left(b^2x+c^2y\right)}{\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]}=$](//latex.artofproblemsolving.com/c/6/5/c65e263239ad3e4e4fca453cf55e7a4916cf3718.png)
![$\frac {4axyS\left(b^2x+c^2y\right)}{\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]}\implies$](//latex.artofproblemsolving.com/1/9/d/19dd5c8108e9ebbd3cf2e3623b4aaada3570b77b.png)
.
Remark. IF
is an -isosceles triangle, THEN ![$2[ABC]=\left(\frac {b^2-a^2}{xy}+\frac {a^2}{b^2}\right)\cdot [QPR]$](//latex.artofproblemsolving.com/e/b/b/ebbdc384c68a483614a41dc0cd1ca0e53d2e66e2.png)
. Observe that ![$a=b=c\implies [QPR]=2[ABC]$](//latex.artofproblemsolving.com/4/3/8/438fa71cf4c9f9be0bf19cfef04606482f215c6e.png)
.
PP17. Let with circumcircle and 
. Let 
. Suppose w.l.o.g. that 
and 
separates , 
. Prove that 
.
Proof 1. Prove easiy
. So 
Proof 2 (own). I"ll use the well-known relation
in the triangles 
. Apply the Ptolemy's theorem in 
.
PP18. Let a trapezoid so that 
, . and 
, 
so that 
and 
. Let
for which 
. Prove that 
.
Proof. Denote
and 
. Is well-known that 
where 
and 
and 
. Observe that
. In conclusion, 
.
PP19 (Ruben Dario). Let
be a triangle with 
and the circle . For a point 
construct 
such
that
, 
so that 
separates 
and . The line 
cut again circumcircle of 
in 
. Prove that the ray 
is bisector of 
.
Proof. Let
and 
. Therefore 
is cyclically and 
.
Thus,
and 
. From 
and 
obtain that 
, i.e. the ray is bisector of .
PP20. Let
be a right angle with 
. Consider the circle 
and the circle 
with the diameter ![$[OA]$](//latex.artofproblemsolving.com/c/8/8/c88c6759afc11d97267e95d4023b4419386ea20f.png)
. Let be a point 
so that the line
doesn't separate and . Construct the rectangle 
where 
and 
. Prove that the ratio 
is constant.
Proof 1. If
, then 
and 
. Using the relation (*) obtain that 
. In conclusion, 
.
Remark. Let an-isosceles with 
so that 
and 
so that 
. Then 
. Indeed, denote 
for which
. Thus, 
and 
and 
are cyclically. Hence 
.
In conclusion,
.
Proof 2. Denote
. Observe that 
and 
is cyclically. Thus, 
.
PP21 (Josue Garcia Piscoya). Let be an acute triangle with the orthocenter . Prove that 
, where 
is the length of the inradius of 
.
Proof. I"ll use relations
. Let so that 
. Thus, 
![$\frac {[BHC]}{[ABC]}=\frac {DH}{DA}=$](//latex.artofproblemsolving.com/9/e/8/9e8a841da9f6fd11f0760fe8e03a73299b091234.png)
![$\boxed{\boxed{[BHC]=S\cot B\cot C}}=$](//latex.artofproblemsolving.com/1/e/2/1e29c453b34b3312bd23209f227437b2a814bbea.png)
![$\boxed{[BHC]=\frac {s^2}4\tan\frac A2\left(1-\tan^2\frac B2\right)\left(1-\tan^2\frac C2\right)}\ (4)$](//latex.artofproblemsolving.com/7/b/f/7bf8e508d48abb31427b2da5f79c64dc28e03038.png)
.
Otherwise.![$[BHC]=\frac 12\cdot HB\cdot HC\cdot\sin\left(\widehat{BHC}\right)=$](//latex.artofproblemsolving.com/2/0/6/206a8f748eab72bafd951e212d0ee2bd87f1f475.png)
![$[BHC]=\frac {s^2}{4}\tan\frac A2\left(1-\tan^2\frac B2\right)\left(1-\tan\frac C2\right)$](//latex.artofproblemsolving.com/1/6/1/161d1474c0e187549c862dc91a9edb23d918d281.png)
.
. Thus, ![$r_1=\frac {2\cdot [BHC]}{HA+HB+BC}\stackrel{(4\wedge 5)}{\implies}$](//latex.artofproblemsolving.com/6/e/a/6ea88567eb63fe212782bcee479789d121181bf8.png)
.
Remark.
.Define analogously inradii 
, 
for 
and 
respectively. Thus,
![$\frac s2\sum\left[1-\left(\tan\frac B2+\tan\frac C2\right)+\tan\frac B2\tan\frac C2\right]\stackrel{(1)}{=}$](//latex.artofproblemsolving.com/a/7/d/a7d0ec7d55c81b8fa1da8a04e50417474ebcd673.png)
.
Remark.
.
-right 
with incircle 
which touches it at 
. Let 
. Prove that 
.Proof 1. Let
and observe that 
. Thus, 
. Denote 
and observe that 
.Proof 2.
.Since 
, obtain that 
![$\boxed{c\left[c^2+(s-a)^2\right]=s(s-b)^2}\ (*)$](http://latex.artofproblemsolving.com/4/5/2/4529f099a82892100ad8635bb38055747f0f19eb.png)
.Denote
. Thus, 
and 
. The relation 
becomes 
. Thus, 
. In conclusion, 
.An easy extension. Let
be a triangle with 
. The its incircle touches it at 
, 
and
. Denote . Prove that 
.Proof. Denote
and 
. Thus, 
. Therefore, 
.PP12 (Elberling Vargas Diaz). Let
be a square with center 
. Denote the midpoints 
, 
, 
of 
, 
, 
respectively. Denote
and 
so that the line 
is tangent to the incircle 
of the given square. Denote 
. Find 
.Proof. Suppose w.l.o.g. that
. Denote 
and 
. Thus, is tangent to 
, i.e.
, i.e. 
. Thus, 
.An easy extension. Let
be a square with center . Denote the midpoints , , and 
of , , and 
respectively. Denote , , 
, 
so that and 
are tangent to the incircle of the given square. Denote 
. Prove that 
.Proof. Suppose w.l.o.g. that
. Denote 
. Therefore,
. Observe that
. In conclusion, 
.PP13. Let two lines
and 
(concurrent or parallel) what are tangent to a circle 
in its points and 
respectively. Prove that
the chain 
.Proof.
.Particular cases.
Let with incircle what touches 
at 
and 
. Prove that 
.
Let with incircle what touches 
at 
. Denote 
. Prove that 
and 
.PP14 (Silver Samuel Palacios). Let
so that 
and 
. Consider 
for which construct the rectangle 
with 
and
. Denote 
, 
so that 
, 
and 
. Prove that 
and 
.Proof. Denote
and 
. Thus, 
. Apply the Menelaus' theorem to:
.
.
.
. Therefore,
and 
. Thus, 
.Apply the Menelaus' theorem to
.PP15. Let convex
with 
. Denote 
. Prove that 
.Proof. Prove easily that
. In conclusion, . Nice and easy problem ! PP16. Let an equilateral
with circumcircle . For 
so that separates and 
denote the 
. Prove that ![$[QPR]=2\cdot [ABC]$](http://latex.artofproblemsolving.com/2/8/2/2822583080d0a905e17868367133dd1b124946e0.png)
.Proof. Suppose w.l.o.g. that
and denote 
where 
. Thus, 
and
![$[ABC]=\frac {\sqrt 3}{4}$](http://latex.artofproblemsolving.com/3/7/2/372282ac1d07c3624ff15b7dda80514c707813db.png)
. Prove easily that 
and 
. Apply Menelaus' theorem to the transvesals: 
.Thus,
![$[QPR]=[QAR]-[QAP]-[RAP]\iff$](http://latex.artofproblemsolving.com/f/6/0/f605c0da15fa1599d8454d3b982cfbe7f63bbf42.png)
![$2\cdot [QPR]=AQ\cdot AR\cdot\sin\widehat{BAC}-AQ\cdot BP\cdot\sin\widehat{ABC}-AR\cdot CP\cdot\sin\widehat{ACB}=$](http://latex.artofproblemsolving.com/f/e/c/fec97ed25fdf43860289695e83ab42588c84f770.png)
![$4\cdot [ABC]\implies$](http://latex.artofproblemsolving.com/e/d/c/edc6e049b4399a29916886408f105673d8df012b.png)
.An easy extension. Let
with the circumcircle . For so that separates and denote so that
doesn't separate 
and 
. Let 
and 
. Prove that ![$[QPR]=\frac {2axy\left(b^2x+c^2y\right)}{\left[b^2x+\left(c^2-a^2\right)y\right]\left[c^2y+\left(b^2-a^2\right)x\right]}\cdot [ABC]$](http://latex.artofproblemsolving.com/6/f/f/6ffb4d6708f6fc947d6093749f2d914c9ea174e8.png)
.Proof. Apply the Stewart's relation to
in 
. Thus, 
and 
. Apply the Menelaus' theorem:
. Therefore, ![$2\cdot [QPR]=AQ\cdot AR\cdot\sin\widehat{BAC}-$](http://latex.artofproblemsolving.com/3/3/4/3340d14adb222416c6506371c6ffdcdf9dff955e.png)
![$\frac{bc\left(b^2x+c^2y\right)^2\sin A}{\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]} -$](http://latex.artofproblemsolving.com/a/c/4/ac4b79510e1109b11b6178cc60428bf4fafd1500.png)
![$\frac {bcx\left(b^2x+c^2y\right)}{2R\left[c^2y+\left(b^2-a^2\right)x\right]}-$](http://latex.artofproblemsolving.com/d/8/4/d840d4d5b41a550970bef00a3a3badf21cfb4e60.png)
![$\frac {bcy\left(b^2x+c^2y\right)}{2R\left[b^2x+\left(c^2-a^2\right)y\right]}=$](http://latex.artofproblemsolving.com/a/5/0/a50fa687882d70e5201d5a82366308c89f0669f6.png)
![$\frac{abc\left(b^2x+c^2y\right)^2-bcx\left(b^2x+c^2y\right)\left[b^2x+\left(c^2-a^2\right)y\right]-bcy\left(b^2x+c^2y\right)\left[c^2y+\left(b^2-a^2\right)x\right]}{2R\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]}=$](http://latex.artofproblemsolving.com/3/c/7/3c7de710d7df08c942bdf13e4350b5078b56c7c2.png)
![$\frac {bc\left(b^2x+c^2y\right)}{2R\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]}\cdot$](http://latex.artofproblemsolving.com/0/6/7/067d6939e960873d559d16f78be7c5a160bb7a38.png)
![$\left\{a\left(b^2x+c^2y\right)-x\left[b^2x+\left(c^2-a^2\right)y\right]-y\left[c^2y+\left(b^2-a^2\right)x\right]\right\}=$](http://latex.artofproblemsolving.com/7/7/0/7705ff4eda9afa1c215f51ad044c4d5e935c77b5.png)
![$\frac {h_a\left(b^2x+c^2y\right)}{\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]}$](http://latex.artofproblemsolving.com/d/c/c/dcc84317853e28b9ee0cac305600b2d138d87bb8.png)
![$\cdot\left[b^2x(a-x)+c^2y(a-y)+xy\left(2a^2-b^2-c^2\right)\right]=$](http://latex.artofproblemsolving.com/f/d/3/fd3f433fd70ed7f6bbeb7d83846fb9ea308b4ac6.png)
![$\frac {2a^2h_axy\left(b^2x+c^2y\right)}{\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]}=$](http://latex.artofproblemsolving.com/c/6/5/c65e263239ad3e4e4fca453cf55e7a4916cf3718.png)
![$\frac {4axyS\left(b^2x+c^2y\right)}{\left[c^2y+\left(b^2-a^2\right)x\right]\left[b^2x+\left(c^2-a^2\right)y\right]}\implies$](http://latex.artofproblemsolving.com/1/9/d/19dd5c8108e9ebbd3cf2e3623b4aaada3570b77b.png)
.Remark. IF
is an -isosceles triangle, THEN ![$2[ABC]=\left(\frac {b^2-a^2}{xy}+\frac {a^2}{b^2}\right)\cdot [QPR]$](http://latex.artofproblemsolving.com/e/b/b/ebbdc384c68a483614a41dc0cd1ca0e53d2e66e2.png)
. Observe that ![$a=b=c\implies [QPR]=2[ABC]$](http://latex.artofproblemsolving.com/4/3/8/438fa71cf4c9f9be0bf19cfef04606482f215c6e.png)
.PP17. Let
with circumcircle and 
. Let 
. Suppose w.l.o.g. that 
and 
separates , 
. Prove that 
.Proof 1. Prove easiy
. So 
Proof 2 (own). I"ll use the well-known relation
in the triangles 
. Apply the Ptolemy's theorem in 
.PP18. Let a trapezoid
so that 
, . and 
, 
so that 
and 
. Let
for which 
. Prove that 
.Proof. Denote
and 
. Is well-known that 
where 
and 
and 
. Observe that
. In conclusion, 
.PP19 (Ruben Dario). Let
be a triangle with 
and the circle . For a point 
construct 
suchthat
, 
so that 
separates 
and . The line 
cut again circumcircle of 
in 
. Prove that the ray 
is bisector of 
.Proof. Let
and 
. Therefore 
is cyclically and 
.Thus,
and 
. From 
and 
obtain that 
, i.e. the ray is bisector of .PP20. Let
be a right angle with 
. Consider the circle 
and the circle 
with the diameter ![$[OA]$](http://latex.artofproblemsolving.com/c/8/8/c88c6759afc11d97267e95d4023b4419386ea20f.png)
. Let be a point 
so that the line
doesn't separate and . Construct the rectangle 
where 
and 
. Prove that the ratio 
is constant.Proof 1. If
, then 
and 
. Using the relation (*) obtain that 
. In conclusion, 
.Remark. Let an
-isosceles with 
so that 
and 
so that 
. Then 
. Indeed, denote 
for which
. Thus, 
and 
and 
are cyclically. Hence 
.In conclusion,
.Proof 2. Denote
. Observe that 
and 
is cyclically. Thus, 
. PP21 (Josue Garcia Piscoya). Let
be an acute triangle with the orthocenter . Prove that 
, where 
is the length of the inradius of 
.Proof. I"ll use relations
. Let so that 
. Thus, 
![$\frac {[BHC]}{[ABC]}=\frac {DH}{DA}=$](http://latex.artofproblemsolving.com/9/e/8/9e8a841da9f6fd11f0760fe8e03a73299b091234.png)
![$\boxed{\boxed{[BHC]=S\cot B\cot C}}=$](http://latex.artofproblemsolving.com/1/e/2/1e29c453b34b3312bd23209f227437b2a814bbea.png)
![$\boxed{[BHC]=\frac {s^2}4\tan\frac A2\left(1-\tan^2\frac B2\right)\left(1-\tan^2\frac C2\right)}\ (4)$](http://latex.artofproblemsolving.com/7/b/f/7bf8e508d48abb31427b2da5f79c64dc28e03038.png)
.Otherwise.
![$[BHC]=\frac 12\cdot HB\cdot HC\cdot\sin\left(\widehat{BHC}\right)=$](http://latex.artofproblemsolving.com/2/0/6/206a8f748eab72bafd951e212d0ee2bd87f1f475.png)
![$[BHC]=\frac {s^2}{4}\tan\frac A2\left(1-\tan^2\frac B2\right)\left(1-\tan\frac C2\right)$](http://latex.artofproblemsolving.com/1/6/1/161d1474c0e187549c862dc91a9edb23d918d281.png)
.
. Thus, ![$r_1=\frac {2\cdot [BHC]}{HA+HB+BC}\stackrel{(4\wedge 5)}{\implies}$](http://latex.artofproblemsolving.com/6/e/a/6ea88567eb63fe212782bcee479789d121181bf8.png)
.Remark.
.Define analogously inradii 
, 
for 
and 
respectively. Thus,
![$\frac s2\sum\left[1-\left(\tan\frac B2+\tan\frac C2\right)+\tan\frac B2\tan\frac C2\right]\stackrel{(1)}{=}$](http://latex.artofproblemsolving.com/a/7/d/a7d0ec7d55c81b8fa1da8a04e50417474ebcd673.png)
.Remark.
.This post has been edited 320 times. Last edited by Virgil Nicula, Feb 11, 2018, 3:16 PM
August 2018
January 2018