392. Art of Problem Solving (geometry).
by Virgil Nicula, Mar 20, 2014, 10:50 PM
PP0. Prove that in 
there is the chain 
(standard notations).
Proof. Are well-known
The product of the inequalities 
, i.e.
a.s.o. 
, i.e. 
from where obtain 
. Applly 
![$ \sqrt[3]{p-a)(p-b)(p-c)}$](//latex.artofproblemsolving.com/7/8/b/78beb285b6aa8fd05b42fb1a35da74648251a9e3.png)
![$ p\ge 3\sqrt[3]{pr^2}$](//latex.artofproblemsolving.com/9/4/8/9487fe71b62bb208ba3731d1f3b3508811bf4d18.png)
In the inequality 
for 
obtain that
![$ 3p(p-a)(p-b)(p-c)\le \left[r(4R+r)\right]^2$](//latex.artofproblemsolving.com/d/e/0/de05838d17b91050bf692b381dac3961a7c64cb1.png)
Since 
get 
, i.e. 
. Otherwise. 
. Add 
and 
.
Remark.
I used the well-known identities 
.
PP1. Let
be an interior point of the parallelogram 
. Let the points 
such that
and 
. Prove that ![$[BGHIDC]=2\cdot [AEF]$](//latex.artofproblemsolving.com/d/4/e/d4ee31483e0a94a455ee1435654d2d63e799fde4.png)
, where ![$[XYZ]$](//latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
is the area of 
.
Proof 1. Observe that![$\boxed{\ [AEF]=[EFH]+[AEH]+[AFH]\ }\ (1)$](//latex.artofproblemsolving.com/8/9/f/89f80199751ebfa1e6b601c7036e1ccb47c43a81.png)
and ![$\boxed{\ [CEHF]=2\cdot [EFH]\ }\ (2)$](//latex.artofproblemsolving.com/4/7/9/4794bcd4f7da3be72c602b660ad1c16326fed0c9.png)
. Thus,
![$\odot\begin{array}{cccccccc}
\nearrow & BA\parallel EH & \implies & [BEH]=[AEH] & \implies & 2\cdot [BEH]=2\cdot [AEH] & \implies & \boxed{\ [BEHG]=2\cdot [AEH]\ }\\\\
\searrow & DA\parallel FH & \implies & [DFH]=[AFH] & \implies & 2\cdot [DFH]=2\cdot [AFH] & \implies & \boxed{\ [DFHI]=2\cdot [AFH]\ }\end{array}\ (3)\implies$](//latex.artofproblemsolving.com/a/1/7/a173a2665e6e4ac565b2558a0932a9bbc18d3b79.png)
![$[BGHIDC]=[CEHF]+[BEHG]+$](//latex.artofproblemsolving.com/7/f/c/7fcbc887101394743b365e158b849640093c3864.png)
![$[DFHI]\ \stackrel{(2\wedge 3)}{=}\ 2\cdot [EFH]+2\cdot [AEH]+2\cdot [AFH]=$](//latex.artofproblemsolving.com/b/0/9/b0906af01b497700105aab6cf8bb30c448a90667.png)
![$2\cdot ([EFH]+[AEH]+[AFH])\stackrel{(1)}{=}\ 2\cdot [AEF]$](//latex.artofproblemsolving.com/4/5/5/455ca3d07b0de2cf3e1e2819c6689acb2b86b50c.png)
.
Proof 2.![$\sigma\equiv [AEF]\ \implies\ [AEF]+[ABE]+[ADF]+[CEF]=[ABCD]\iff$](//latex.artofproblemsolving.com/6/e/c/6eca9a7387e88afacb8a35cad36827fcb0fbeb70.png)
![$2\sigma +[ABEI]+[ADFG]+[CEHF]=2[ABCD]\iff$](//latex.artofproblemsolving.com/d/f/6/df60576b7217cdbb09a6ca7c5652ed3deed35bb7.png)
![$2\sigma +[ABEI]+[ADFG]+[CEHF]=$](//latex.artofproblemsolving.com/c/b/9/cb979984528f5dd7e4843b8098fcbf2978647f90.png)
![$[ABCD]+([ABEI]+[CEHF]+[DIHF])\iff$](//latex.artofproblemsolving.com/4/d/8/4d84bf517518a3ef12bc05e460ea5985e4568855.png)
![$2\sigma +[ADFG]=[ABCD]+[DIHF]\iff$](//latex.artofproblemsolving.com/b/5/0/b50678ff2b585e04998dd672b997425c198e55c4.png)
![$2\sigma =[BCFG]+[DIHF]\iff$](//latex.artofproblemsolving.com/f/c/0/fc0e3fa94111e8e707fcc6aef99b69cf74e4baa5.png)
![$2\sigma =[BGHIDC]\iff$](//latex.artofproblemsolving.com/d/8/2/d82c35710a15dac8eb3a163e9444cbf7622ab396.png)
.
PP2. Prove that in any triangle
exists the equivalence 
(standard notations).
Proof 1. Let
and 
, i.e. 
.
Proof 2. Construct
so that 
. Thus, 
. Therefore,
.
PP3. Prove the Heron's formula (with more proofs)
the area 
of the triangle is given by 
, where 
(standard notations).
Proof 1 (classic). Denote the projection
of 
on 
. Suppose w.l.o.g. that 
. Thus, 
![$16S^2=\left[(a+c)^2-b^2\right]\left[b^2-(a-c)^2\right]\implies$](//latex.artofproblemsolving.com/e/f/2/ef2be6da9fd53b9a0a6e08d5b0672debc1e98182.png)
.
Remark. If denote
, then 
and the formula 
becomes 
.
Thus,
and 
I used the relations 
.
Proof 2. Is well-known that
. Let 
, where 
is the incircle and 
is the -excircle. So 
and
. In conclusion, 
.
Proof 3. Is well-known that
. Let 
, where 
is the 
-excircle and 
is the 
-excircle. Thus,
and 
.
Proof 4.
a.s.o. 
![$16S^2=\sum\left[a^4-\left(b^2-c^2\right)^2\right]\iff$](//latex.artofproblemsolving.com/b/3/9/b3953df333e16f332eb5a6fb52d7a7a51032581b.png)
.
Proof 5. Let the orthocenter and 
. Suppose that is acute. Then ![$2S=\sum 2[BHC]=\sum aHD=$](//latex.artofproblemsolving.com/4/5/8/458b9c4d7482cdd8b145707c4d84c737c030d33e.png)
Remark.
Otherwise, ![$S=\sum [BOC]=\frac R2\cdot \sum a\cos A\implies \sum a\cos A=\frac {2S}{R}$](//latex.artofproblemsolving.com/1/1/3/1131aa3c97144e7294d64def82e3f26ea32f10a3.png)
, where 
is the length of the circumradius.
PP4. Prove that in any triangle there is the identity 
, where (standard notations). See and here.
Proof 1. Prove easily that
.
Indeed, let
Proof 2. Let incircle , -excircle , circumcircle 
and 
. Thus,
and 
. So
. Thus, 
. The sum of 
.
Proof 3.
. Hence 
i.e.
, where 
.
Proof 4. Let incircle circumcircle . From 
obtain 
from 
, where 
The relation
is obtained from the Ptolemy's theorem in the quadrilateral 
, where 
. Am folosit 
.
Thus obtain
which is an interesting identity. Hence 
and
.
Obtain 
.
Remark. Can obtain without the Ptolemy's theorem let
and from 
and 
obtain
, adica relatia 
a.s.o. Tot asa 
![$s^2=\sum\left[a^2-(s-a)^2\right]\Longleftrightarrow$](//latex.artofproblemsolving.com/d/b/2/db29b55a8def1933ca3e4273a2872bc4634fb74f.png)
and O.K.
PP5.Prove that in exists Euler's relation 
.
Proof 1 (own). Let the midpoint
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the diameter ![$[NS]$](//latex.artofproblemsolving.com/8/0/d/80de18ed66832d94ce5323f567fc5b59dc992316.png)
of the circumcircle 
, where 
and separates , . Thus, the projection of ![$[OI]$](//latex.artofproblemsolving.com/1/e/6/1e6d88a33008d2bd15b8d0f7de4ff31d77c52d0f.png)
on is
equally to
and the projection of on 
is equally to 
, where 
. Thus, 
. Thus, 
![$\frac {(s-b)(s-c)[(s-a)+a]}{s}-(s-b)(s-c)=$](//latex.artofproblemsolving.com/1/f/d/1fd6b1e150deab25628e2e6fa0b3b672ba93581c.png)
. In conclusion, 
.
Proof 2 (classic). Let the midpoint of , the tangent point 
of incircle 
and diameter of circumcircle , where
and separates , . Thus, 
and 
.
Proof 3 (classic). Is well-known that
. Hence 
![$\frac {2Rr[(b+c-a)+a]}{b+c}=2Rr\implies$](//latex.artofproblemsolving.com/e/e/7/ee71374923d378847051b168dd9129add4ba9919.png)
.
Proof 4. Apply the generalized Pythagoras' relation in
.
PP6. Let
be a right circular cone with radius 
and generatrix 
Prove that
if its volume is constant, then: its lateral surface is minimum iff
; its total surface is minimum iff 
.
Proof. Denote the circle
from the base and the apex with 
. Thus, 
and volume 
is constant, i.e.
is constant. Lateral surface area is 
and total surface area is 
.
is constant 
is constant 
is constant . Thus, 
is 
is
is 
is 
is 
.
is constant 
is constant 
is constant 
is constant . Therefore,
is 
is 
is 
.
PP7. Prove that for exists Weitzenböck's inequality 
.
Proof. Prove easily
.
Method
Since 
we can apply Tchebyshev's
inequality
.
Method
I"ll apply inequality 
, where 
.
PP8.
Proof 1. Let
. Thus, 
and 
.
.
Prove easily that 
(
is circumcenter of 
)
and
. Thus, 
.
Proof 2 (metric).
Hence 
.
Remark. Exist relations
.
Indeed,
. Since 
.
Otherwise.
.
Remark.
PP9. Let with incenter and 
. Prove that 
.
Proof 1. Apply the theorem of Sines in
.Thus, 
. Define over 
an equivalence relation
, i.e. 
and 
have same signature. Observe that 
and 
. Hence 
, i.e. 
.
PP10. Let with circumradius 
and inradius . Prove that 
.
Proof.
and 
. Suppose that 
and 
. Thus,
or 
and 
.
Contraexample.
, where 
and 
.
Remark.
. In this case 
PP11. Let with the circumcircle 
, incircle 
-excircle 
and
so that 
and 
. Prove that 
.
Proof.
.
Thus,
. From the -right 
obtain that 
.
Otherwise.
.
PP12. Let and 
fot which 
and 
, where 
. Prove that Stewart's relation 
.
Proof. I"ll use property:
. Let 
and suppose w.l.o.g. 
, where 
.
Thus,
.
PP13. Prove that a plane which pass through midpoints of two opposite edges of a tetrahedron cut the tetrahedron in two solids with same volume.
Proof. Let midpoints
of ![$[AB],$](//latex.artofproblemsolving.com/5/8/e/58e377d741d6612acac694994b20fa626c18e0ce.png)
![$[CD]$](//latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
respectively. Suppose the plane 
cut in 
and ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
in 
. Appear two cases:
are midpoints of , respectively and 
. Let midpoint of ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and volume 
of
. Observe that 
is a prism. Its volum is ![$v=v[AMPCNQ]=$](//latex.artofproblemsolving.com/7/0/f/70fb027d7fed244082056a7abed425092d3dad20.png)
![$v[AMPRNQ]+v[RPCN]\implies$](//latex.artofproblemsolving.com/b/2/c/b2c745bde7430f378eddc1b35c9ab8592d5b4ad4.png)
![$\frac vV=\frac {v[AMPRMQ]}V+$](//latex.artofproblemsolving.com/b/6/9/b696857565332f6163670e17adfe67defba642ec.png)
![$\frac {v[RPCN]}V=$](//latex.artofproblemsolving.com/b/0/5/b05220d8de12b3d22937165aeed8bca94c1e1817.png)
![$\frac {[AMQ]\cdot \delta_{(ABD)}(R)}{[ABD]\cdot \delta _{(ABD)}(C)}+$](//latex.artofproblemsolving.com/5/1/6/5160d7e833a241bb2e34aad37cb5214081d5da9f.png)
![$\frac {[PCN]\cdot\delta_{(PCN)}(R)}{[BCD]\cdot\delta_{(BCD)}(A)}=$](//latex.artofproblemsolving.com/9/5/e/95ebf925c21fad0fe19b161b0a46914d6ee54be0.png)
![$\frac {[AMQ]}{[ABD]}\cdot \frac {RA}{CA}+$](//latex.artofproblemsolving.com/e/2/2/e2291fa139f17e200ec26d129173cadffe2e39f1.png)
![$\frac {[PCN]}{[BCD]}\cdot \frac {RC}{AC}=$](//latex.artofproblemsolving.com/2/b/5/2b533befae8287266532035531cb2f7b6c6b69cf.png)
.
. Let 
and with Menelaus' theorem prove easily 
. suppose w.l.o.g.
, adica 
si 
. Required volume is ![$v[RMAQ]-v[RPCN]$](//latex.artofproblemsolving.com/7/0/b/70bc54866933d1653a48f7bd3888326a359286b6.png)
and ![$\frac vV=\frac {v[RMAQ]}V-\frac {v[RPCN]}V=$](//latex.artofproblemsolving.com/a/1/0/a109199220d21d96a8d9a3232e5a7dc0a51d65fc.png)
![$\frac {[AMQ]\cdot\delta_{(ABD)}(R)}{[ABD]\cdot\delta_{(ABD)}(C)}-\frac {[PCN]\cdot\delta_{(BCD)}(R)}{[BCD]\cdot\delta_{(BCD)}(A)}=$](//latex.artofproblemsolving.com/3/2/4/324677002f24f54bcb8faf02bf8b41f35d1c4437.png)
.
PP14. Let with incenter , centroid 
and orthocentre . Prove that 
.
Proof. Choose the origin of the vectorial system. Thus, 
, 
and 
, 
. We know that 
,
and 
. Apply the theorem of - median in 
. Thus, 
. So, 
![$\frac{1}{3}\left[(R-2r)^{2}-(R^{2}-2Rr)\right]=$](//latex.artofproblemsolving.com/4/e/8/4e896d6270aa63a276989234b0d575795c30a020.png)
.
Thus
, i.e. 
PP15. Let with incircle . Prove that 
.
Proof. Let
so that 
. Prove easily that 
, i.e. 
is an -right isosceles and
is the circumcenter of . Thus, 
and 
. Shortly, 
.
Remark.
there is the chain of the equalities 
.
PP16. The point is a midpoint of 
. Circle 
which passes through and intersect circle 
which passes through and in two different points and .
Point is midpoint of arc 
of circle which doesn't contain . Point is a midpoint of arc 
of circle which doesn't contain . Prove that 
.
Here is a short metrical proof.
Lemma. Let with circumcircle 
and middlepoint 
of the arc 
which doesn't contain vertex . Then exists relation 
Proof 1. Let
. Thus, 
and 
.
Thus,
Therefore,
Proof 2. In have 
and 
. Thus, ![$4R^2\left[\sin^2\left(B+\frac A2\right)-\sin^2\frac A2\right]=$](//latex.artofproblemsolving.com/5/b/d/5bd74212f4cb1af130a164224f6012f666508811.png)
Remark. If denote the middlepoint
of the arc which contains the vertex , then there is and the relation 
because 
In the proof of the lemma I used the relation 
and theorem of the Sinus.
Proof 3. Let
and incircle . Hence 
Remark.. I denoted the power
of w.r.t. circumcircle and I used the relations: 
,
, 
, 
(from 
), 
and 
Proof 4.
![$\frac{bc\left[(b+c)^2-a^2\right]}{4p(p-a)}=bc\ .$](//latex.artofproblemsolving.com/8/3/b/83be30acf951a6267879504a29b376f731df0b4a.png)
Remark. I used the relations, 
and 
Proof of the proposed problem. Apply the above lemma for
and 
and
. But 
. Thus, 
PP17 (Gustavo Jimmy Garcia Paytan). Let-right with -excircle 
and -excircle 
incircles 
and
of 
and 
respectively. Let 
of 
and 
. Prove that 
.
Proof.
and 
are cyclic and 
, i.e. 
Since 
and
The relations and implies relation 
Remark. Otherwise can apply the well-known identities in
.
PP18. Let and an its interior so that 
and 
. Prove that 
and 
.
Proof. Let
and 
. So 
. Thus, 
and 
. Hence 
is equilateral and .
PP19. Let and for 
let the cevian 
, where 
. Prove that ![$\frac 1{[BDF]}+\frac 2{[DEF]}=\frac 1{[DEFP]}+\frac 2{[BDPF]}$](//latex.artofproblemsolving.com/3/3/2/332d25aa9c54c1df9a36529a4ffdaf5467d7782b.png)
.
Proof. Let
Is well-known that the division 
is harmonically, i.e. 
and 
.
Let![$\left\{\begin{array}{ccc}
b & = & [BDF]\\\\
m & = & [DEF]\\\\
s & = & [DFP]\end{array}\right\|$](//latex.artofproblemsolving.com/3/a/e/3ae190f2d7678ccd43ee4be45b6e08543736d404.png)
. Thus, ![$\frac 1{[BDF]}+\frac 2{[DEF]}=\frac 1{[DEFP]}+\frac 2{[BDPF]}\iff$](//latex.artofproblemsolving.com/8/d/d/8ddc5d24b779e5a96769b24a039f73324ccae40e.png)
![$\frac {[BDF]}{[DFP]}=1+2\cdot \frac {[BDF]}{[DEF]}\iff$](//latex.artofproblemsolving.com/b/e/1/be1134fa290153ccee63db458a648fc01d2e540d.png)
what is the relation . Otherwise. 
what is the relation . See here
there is the chain 
(standard notations).Proof. Are well-known
The product of the inequalities 
, i.e.
a.s.o. 
, i.e. 
from where obtain 
. Applly 
![$ \sqrt[3]{p-a)(p-b)(p-c)}$](http://latex.artofproblemsolving.com/7/8/b/78beb285b6aa8fd05b42fb1a35da74648251a9e3.png)
![$ p\ge 3\sqrt[3]{pr^2}$](http://latex.artofproblemsolving.com/9/4/8/9487fe71b62bb208ba3731d1f3b3508811bf4d18.png)
In the inequality 
for 
obtain that![$ 3p(p-a)(p-b)(p-c)\le \left[r(4R+r)\right]^2$](http://latex.artofproblemsolving.com/d/e/0/de05838d17b91050bf692b381dac3961a7c64cb1.png)
Since 
get 
, i.e. 
. Otherwise. 
. Add 
and 
.Remark.
I used the well-known identities 
.PP1. Let
be an interior point of the parallelogram 
. Let the points 
such that
and 
. Prove that ![$[BGHIDC]=2\cdot [AEF]$](http://latex.artofproblemsolving.com/d/4/e/d4ee31483e0a94a455ee1435654d2d63e799fde4.png)
, where ![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
is the area of 
.Proof 1. Observe that
![$\boxed{\ [AEF]=[EFH]+[AEH]+[AFH]\ }\ (1)$](http://latex.artofproblemsolving.com/8/9/f/89f80199751ebfa1e6b601c7036e1ccb47c43a81.png)
and ![$\boxed{\ [CEHF]=2\cdot [EFH]\ }\ (2)$](http://latex.artofproblemsolving.com/4/7/9/4794bcd4f7da3be72c602b660ad1c16326fed0c9.png)
. Thus,![$\odot\begin{array}{cccccccc}
\nearrow & BA\parallel EH & \implies & [BEH]=[AEH] & \implies & 2\cdot [BEH]=2\cdot [AEH] & \implies & \boxed{\ [BEHG]=2\cdot [AEH]\ }\\\\
\searrow & DA\parallel FH & \implies & [DFH]=[AFH] & \implies & 2\cdot [DFH]=2\cdot [AFH] & \implies & \boxed{\ [DFHI]=2\cdot [AFH]\ }\end{array}\ (3)\implies$](http://latex.artofproblemsolving.com/a/1/7/a173a2665e6e4ac565b2558a0932a9bbc18d3b79.png)
![$[BGHIDC]=[CEHF]+[BEHG]+$](http://latex.artofproblemsolving.com/7/f/c/7fcbc887101394743b365e158b849640093c3864.png)
![$[DFHI]\ \stackrel{(2\wedge 3)}{=}\ 2\cdot [EFH]+2\cdot [AEH]+2\cdot [AFH]=$](http://latex.artofproblemsolving.com/b/0/9/b0906af01b497700105aab6cf8bb30c448a90667.png)
![$2\cdot ([EFH]+[AEH]+[AFH])\stackrel{(1)}{=}\ 2\cdot [AEF]$](http://latex.artofproblemsolving.com/4/5/5/455ca3d07b0de2cf3e1e2819c6689acb2b86b50c.png)
.Proof 2.
![$\sigma\equiv [AEF]\ \implies\ [AEF]+[ABE]+[ADF]+[CEF]=[ABCD]\iff$](http://latex.artofproblemsolving.com/6/e/c/6eca9a7387e88afacb8a35cad36827fcb0fbeb70.png)
![$2\sigma +[ABEI]+[ADFG]+[CEHF]=2[ABCD]\iff$](http://latex.artofproblemsolving.com/d/f/6/df60576b7217cdbb09a6ca7c5652ed3deed35bb7.png)
![$2\sigma +[ABEI]+[ADFG]+[CEHF]=$](http://latex.artofproblemsolving.com/c/b/9/cb979984528f5dd7e4843b8098fcbf2978647f90.png)
![$[ABCD]+([ABEI]+[CEHF]+[DIHF])\iff$](http://latex.artofproblemsolving.com/4/d/8/4d84bf517518a3ef12bc05e460ea5985e4568855.png)
![$2\sigma +[ADFG]=[ABCD]+[DIHF]\iff$](http://latex.artofproblemsolving.com/b/5/0/b50678ff2b585e04998dd672b997425c198e55c4.png)
![$2\sigma =[BCFG]+[DIHF]\iff$](http://latex.artofproblemsolving.com/f/c/0/fc0e3fa94111e8e707fcc6aef99b69cf74e4baa5.png)
![$2\sigma =[BGHIDC]\iff$](http://latex.artofproblemsolving.com/d/8/2/d82c35710a15dac8eb3a163e9444cbf7622ab396.png)
.PP2. Prove that in any triangle
exists the equivalence 
(standard notations).Proof 1. Let
and 
, i.e. 
.Proof 2. Construct
so that 
. Thus, 
. Therefore,
.PP3. Prove the Heron's formula (with more proofs)
the area 
of the triangle is given by 
, where 
(standard notations).Proof 1 (classic). Denote the projection
of 
on 
. Suppose w.l.o.g. that 
. Thus, 
![$16S^2=\left[(a+c)^2-b^2\right]\left[b^2-(a-c)^2\right]\implies$](http://latex.artofproblemsolving.com/e/f/2/ef2be6da9fd53b9a0a6e08d5b0672debc1e98182.png)
.Remark. If denote
, then 
and the formula 
becomes 
.Thus,
and 
I used the relations 
.Proof 2. Is well-known that
. Let 
, where 
is the incircle and 
is the -excircle. So 
and
. In conclusion, 
.Proof 3. Is well-known that
. Let 
, where 
is the 
-excircle and 
is the 
-excircle. Thus,
and 
.Proof 4.
a.s.o. 
![$16S^2=\sum\left[a^4-\left(b^2-c^2\right)^2\right]\iff$](http://latex.artofproblemsolving.com/b/3/9/b3953df333e16f332eb5a6fb52d7a7a51032581b.png)
.Proof 5. Let the orthocenter
and 
. Suppose that is acute. Then ![$2S=\sum 2[BHC]=\sum aHD=$](http://latex.artofproblemsolving.com/4/5/8/458b9c4d7482cdd8b145707c4d84c737c030d33e.png)
Remark.
Otherwise, ![$S=\sum [BOC]=\frac R2\cdot \sum a\cos A\implies \sum a\cos A=\frac {2S}{R}$](http://latex.artofproblemsolving.com/1/1/3/1131aa3c97144e7294d64def82e3f26ea32f10a3.png)
, where 
is the length of the circumradius. PP4. Prove that in any triangle
there is the identity 
, where (standard notations). See and here.Proof 1. Prove easily that
.Indeed, let
Proof 2. Let incircle
, -excircle , circumcircle 
and 
. Thus,
and 
. So
. Thus, 
. The sum of 
.Proof 3.
. Hence 
i.e.
, where 
.Proof 4. Let incircle
circumcircle . From 
obtain 
from 
, where 
The relation
is obtained from the Ptolemy's theorem in the quadrilateral 
, where 
. Am folosit 
.Thus obtain
which is an interesting identity. Hence 
and
.Obtain
.Remark. Can obtain
without the Ptolemy's theorem let
and from 
and 
obtain
, adica relatia 
a.s.o. Tot asa 
![$s^2=\sum\left[a^2-(s-a)^2\right]\Longleftrightarrow$](http://latex.artofproblemsolving.com/d/b/2/db29b55a8def1933ca3e4273a2872bc4634fb74f.png)
and O.K.PP5.Prove that in
exists Euler's relation 
. Proof 1 (own). Let the midpoint
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the diameter ![$[NS]$](http://latex.artofproblemsolving.com/8/0/d/80de18ed66832d94ce5323f567fc5b59dc992316.png)
of the circumcircle 
, where 
and separates , . Thus, the projection of ![$[OI]$](http://latex.artofproblemsolving.com/1/e/6/1e6d88a33008d2bd15b8d0f7de4ff31d77c52d0f.png)
on isequally to
and the projection of on 
is equally to 
, where 
. Thus, 
. Thus, 
![$\frac {(s-b)(s-c)[(s-a)+a]}{s}-(s-b)(s-c)=$](http://latex.artofproblemsolving.com/1/f/d/1fd6b1e150deab25628e2e6fa0b3b672ba93581c.png)
. In conclusion, 
.Proof 2 (classic). Let the midpoint
of , the tangent point 
of incircle 
and diameter of circumcircle , whereand separates , . Thus, 
and 
.Proof 3 (classic). Is well-known that
. Hence 
![$\frac {2Rr[(b+c-a)+a]}{b+c}=2Rr\implies$](http://latex.artofproblemsolving.com/e/e/7/ee71374923d378847051b168dd9129add4ba9919.png)
.Proof 4. Apply the generalized Pythagoras' relation in
.PP6. Let
be a right circular cone with radius 
and generatrix 
Prove thatif its volume is constant, then: its lateral surface is minimum iff
; its total surface is minimum iff 
.Proof. Denote the circle
from the base and the apex with 
. Thus, 
and volume 
is constant, i.e.
is constant. Lateral surface area is 
and total surface area is 
.
is constant 
is constant 
is constant . Thus, 
is 
is 
is 
is 
is 
. is constant 
is constant 
is constant 
is constant . Therefore, is 
is 
is 
.PP7. Prove that for
exists Weitzenböck's inequality 
.Proof. Prove easily
.Method
Since 
we can apply Tchebyshev'sinequality
.Method
I"ll apply inequality 
, where 
.PP8.
Proof 1. Let
. Thus, 
and 
.
.
Prove easily that 
(
is circumcenter of 
)and
. Thus, 
. Proof 2 (metric).
Hence 
.Remark. Exist relations
.Indeed,
. Since 
.Otherwise.
.Remark.
PP9. Let
with incenter and 
. Prove that 
.Proof 1. Apply the theorem of Sines in
.Thus, 
. Define over 
an equivalence relation
, i.e. 
and 
have same signature. Observe that 
and 
. Hence 
, i.e. 
.PP10. Let
with circumradius 
and inradius . Prove that 
.Proof.
and 
. Suppose that 
and 
. Thus,
or 
and 
.Contraexample.
, where 
and 
.Remark.
. In this case 
PP11. Let
with the circumcircle 
, incircle 
-excircle 
and
so that 
and 
. Prove that 
.Proof.
.Thus,
. From the -right 
obtain that 
.Otherwise.
.PP12. Let
and 
fot which 
and 
, where 
. Prove that Stewart's relation 
.Proof. I"ll use property:
. Let 
and suppose w.l.o.g. 
, where 
.Thus,
.PP13. Prove that a plane which pass through midpoints of two opposite edges of a tetrahedron cut the tetrahedron in two solids with same volume.
Proof. Let midpoints
of ![$[AB],$](http://latex.artofproblemsolving.com/5/8/e/58e377d741d6612acac694994b20fa626c18e0ce.png)
![$[CD]$](http://latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
respectively. Suppose the plane 
cut in 
and ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
in 
. Appear two cases:
are midpoints of , respectively and 
. Let midpoint of ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
and volume 
of . Observe that 
is a prism. Its volum is ![$v=v[AMPCNQ]=$](http://latex.artofproblemsolving.com/7/0/f/70fb027d7fed244082056a7abed425092d3dad20.png)
![$v[AMPRNQ]+v[RPCN]\implies$](http://latex.artofproblemsolving.com/b/2/c/b2c745bde7430f378eddc1b35c9ab8592d5b4ad4.png)
![$\frac vV=\frac {v[AMPRMQ]}V+$](http://latex.artofproblemsolving.com/b/6/9/b696857565332f6163670e17adfe67defba642ec.png)
![$\frac {v[RPCN]}V=$](http://latex.artofproblemsolving.com/b/0/5/b05220d8de12b3d22937165aeed8bca94c1e1817.png)
![$\frac {[AMQ]\cdot \delta_{(ABD)}(R)}{[ABD]\cdot \delta _{(ABD)}(C)}+$](http://latex.artofproblemsolving.com/5/1/6/5160d7e833a241bb2e34aad37cb5214081d5da9f.png)
![$\frac {[PCN]\cdot\delta_{(PCN)}(R)}{[BCD]\cdot\delta_{(BCD)}(A)}=$](http://latex.artofproblemsolving.com/9/5/e/95ebf925c21fad0fe19b161b0a46914d6ee54be0.png)
![$\frac {[AMQ]}{[ABD]}\cdot \frac {RA}{CA}+$](http://latex.artofproblemsolving.com/e/2/2/e2291fa139f17e200ec26d129173cadffe2e39f1.png)
![$\frac {[PCN]}{[BCD]}\cdot \frac {RC}{AC}=$](http://latex.artofproblemsolving.com/2/b/5/2b533befae8287266532035531cb2f7b6c6b69cf.png)
.
. Let 
and with Menelaus' theorem prove easily 
. suppose w.l.o.g.
, adica 
si 
. Required volume is ![$v[RMAQ]-v[RPCN]$](http://latex.artofproblemsolving.com/7/0/b/70bc54866933d1653a48f7bd3888326a359286b6.png)
and ![$\frac vV=\frac {v[RMAQ]}V-\frac {v[RPCN]}V=$](http://latex.artofproblemsolving.com/a/1/0/a109199220d21d96a8d9a3232e5a7dc0a51d65fc.png)
![$\frac {[AMQ]\cdot\delta_{(ABD)}(R)}{[ABD]\cdot\delta_{(ABD)}(C)}-\frac {[PCN]\cdot\delta_{(BCD)}(R)}{[BCD]\cdot\delta_{(BCD)}(A)}=$](http://latex.artofproblemsolving.com/3/2/4/324677002f24f54bcb8faf02bf8b41f35d1c4437.png)
.PP14. Let
with incenter , centroid 
and orthocentre . Prove that 
.Proof. Choose the origin
of the vectorial system. Thus, 
, 
and 
, 
. We know that 
,
and 
. Apply the theorem of - median in 
. Thus, 
. So, 
![$\frac{1}{3}\left[(R-2r)^{2}-(R^{2}-2Rr)\right]=$](http://latex.artofproblemsolving.com/4/e/8/4e896d6270aa63a276989234b0d575795c30a020.png)
.Thus
, i.e. 
PP15. Let
with incircle . Prove that 
.Proof. Let
so that 
. Prove easily that 
, i.e. 
is an -right isosceles and is the circumcenter of . Thus, 
and 
. Shortly, 
.Remark.
there is the chain of the equalities 
.PP16. The point
is a midpoint of 
. Circle 
which passes through and intersect circle 
which passes through and in two different points and .Point
is midpoint of arc 
of circle which doesn't contain . Point is a midpoint of arc 
of circle which doesn't contain . Prove that 
.Here is a short metrical proof.
Lemma. Let
with circumcircle 
and middlepoint 
of the arc 
which doesn't contain vertex . Then exists relation 
Proof 1. Let
. Thus, 
and 
.Thus,
Therefore, Proof 2. In
have 
and 
. Thus, ![$4R^2\left[\sin^2\left(B+\frac A2\right)-\sin^2\frac A2\right]=$](http://latex.artofproblemsolving.com/5/b/d/5bd74212f4cb1af130a164224f6012f666508811.png)
Remark. If denote the middlepoint
of the arc which contains the vertex , then there is and the relation 
because 
In the proof of the lemma I used the relation 
and theorem of the Sinus.Proof 3. Let
and incircle . Hence 
Remark.. I denoted the power
of w.r.t. circumcircle and I used the relations: 
,
, 
, 
(from 
), 
and 
Proof 4.
![$\frac{bc\left[(b+c)^2-a^2\right]}{4p(p-a)}=bc\ .$](http://latex.artofproblemsolving.com/8/3/b/83be30acf951a6267879504a29b376f731df0b4a.png)
Remark. I used the relations
, 
and 
Proof of the proposed problem. Apply the above lemma for
and 
and
. But 
. Thus, 
PP17 (Gustavo Jimmy Garcia Paytan). Let
-right with -excircle 
and -excircle 
incircles 
and
of 
and 
respectively. Let 
of 
and 
. Prove that 
.Proof.
and 
are cyclic and 
, i.e. 
Since 
and
The relations and implies relation 
Remark. Otherwise can apply the well-known identities in
.PP18. Let
and an its interior so that 
and 
. Prove that 
and 
.Proof. Let
and 
. So 
. Thus, 
and 
. Hence 
is equilateral and . PP19. Let
and for 
let the cevian 
, where 
. Prove that ![$\frac 1{[BDF]}+\frac 2{[DEF]}=\frac 1{[DEFP]}+\frac 2{[BDPF]}$](http://latex.artofproblemsolving.com/3/3/2/332d25aa9c54c1df9a36529a4ffdaf5467d7782b.png)
.Proof. Let
Is well-known that the division 
is harmonically, i.e. 
and 
.Let
![$\left\{\begin{array}{ccc}
b & = & [BDF]\\\\
m & = & [DEF]\\\\
s & = & [DFP]\end{array}\right\|$](http://latex.artofproblemsolving.com/3/a/e/3ae190f2d7678ccd43ee4be45b6e08543736d404.png)
. Thus, ![$\frac 1{[BDF]}+\frac 2{[DEF]}=\frac 1{[DEFP]}+\frac 2{[BDPF]}\iff$](http://latex.artofproblemsolving.com/8/d/d/8ddc5d24b779e5a96769b24a039f73324ccae40e.png)
![$\frac {[BDF]}{[DFP]}=1+2\cdot \frac {[BDF]}{[DEF]}\iff$](http://latex.artofproblemsolving.com/b/e/1/be1134fa290153ccee63db458a648fc01d2e540d.png)
what is the relation . Otherwise. 
what is the relation . See hereThis post has been edited 284 times. Last edited by Virgil Nicula, Jun 28, 2017, 5:39 AM
August 2018
January 2018