371. Some problems from NMO (final stage) - 2013, Romania.
by Virgil Nicula, Apr 8, 2013, 12:29 AM
Lemma 1. Let 
for which the projection 
of 
on 
belongs to the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Then 
.
Proof. Apply the generalized Pythagoras' relation
. Otherwise. 
.
Can show shorter with trigonometry (the Cosinus' theorem). Indeed,
.
Remark. Using relation
can obtain Heron's formula. Indeed, 
![$\left[2ac+\left(a^2+c^2-b^2\right)\right]$](//latex.artofproblemsolving.com/5/3/9/5398d45be7c21ab05a2b188797f95d8c004857ce.png)
![$\left[2ac-\left(a^2+c^2-b^2\right)\right]=$](//latex.artofproblemsolving.com/b/f/d/bfdfc4fc848dbcc5988195755ed76fe41571f51a.png)
![$\left[(a+c)^2-b^2\right]$](//latex.artofproblemsolving.com/e/d/f/edf1e55b53d83284d19bbabdf0b8835e1f45ea95.png)
![$\left[b^2-(a-c)^2\right]\implies$](//latex.artofproblemsolving.com/5/a/0/5a0ddbd333ec541f42121e09b0bbc495b1319979.png)
.
Remark. Let![$[AL]$](//latex.artofproblemsolving.com/7/3/7/737a9b49658cbe239832952bb575c8bbdf7dbb35.png)
be a diameter of the circumcircle 
. Thus, 
. in conclusion, 
. Hence 
.
PP1 (grade 7 - modified enunciation). Let an-right with 
. Denote 
so that 
,
and 
. Prove that 
and in this case find the ratios 
and 
.
Proof 1. Show easily that
and 
. Apply the upper lemma: 
. Obtain analogously that 
.
Method 1.1. I"ll use the property
. Hence 
. Using substitution 
, relation 
becomes 
Method 1.2. Denote
. Apply the Menelaus' theorem to the transversals 
in the triangles 
respectively:
.
.
Therefore, 
.
Suppose that
. In this case 
. Apply the property 
. Apply the Menelaus' theorem to
and 
. Thus, 
.
I"ll show that
. Indeed, apply the Menelaus' theorem to 
. Therefore, 
. See here.
Proof 2 (analytic). Let
. Thus, 
. So
Lemma 2. Let and 
. Then there is the relation 
.
Proof 3.. Let
. Thus, 
, 
are cyclic, i.e. 
and 
. Apply
lemma 2 and get
. I"ll use method 1.1 
.
PP2 (grade 7). Let with 
and 
and 
so that 
. Prove that 
.
Proof.
belong to the 
-median 
.Thus, 
. Hence . This problem is very easy.
PP3 (grade 8). Ascertain
and 
so that ![$[x] +\left\{\frac 1x\right\}=1.005\cdot n$](//latex.artofproblemsolving.com/9/8/a/98a5c72fe761f445af22b5a25ef0d6591f6fc18a.png)
.
Proof. Show easily that
. Hence 
, i.e. 
. Denote 
, where 
and 
. Thus, the equation becomes
![$[x] +\left\{\frac 1x\right\}=\left(1+\frac 1{200}\right)\cdot n\iff$](//latex.artofproblemsolving.com/1/1/7/1174fac7d1204b66f0ce59275aa8574fda4fff80.png)
![$[x] +\left\{\frac 1x\right\}=n+q+\frac r{200}$](//latex.artofproblemsolving.com/b/f/5/bf50fb40eec6e67a0ce50f5807f9531a71859077.png)
. Hence ![$[x]=n+q$](//latex.artofproblemsolving.com/8/c/a/8cadaca3761d3b8d94d27352827834251b79211b.png)
, i.e. ![$[x]=201q+r$](//latex.artofproblemsolving.com/4/b/b/4bb540e825b667fb493a8ba8e5412ff726f8307a.png)
and 
. Thus, 
, i.e.
and 
. From ![$[x]\le x<[x]+1$](//latex.artofproblemsolving.com/c/8/8/c88cb23bfa4209b9fd8de2b154a3c5402ea5d268.png)
obtain that 
.
Since and 
obtain that 
and 
and 
.
PP4 (grade 9). Let an interior point of for which denote 
. and the radius-length 
of the incircle for 
. Prove that ![$[DEF]\le \frac S4$](//latex.artofproblemsolving.com/6/5/e/65ee17f34b37af3b87ba240cb12f33350a66b3b6.png)
and 
.
Proof. Show easily that
, where 
![$[DEF]=\frac {2S}{(x+1)(y+1)(z+1)}\le$](//latex.artofproblemsolving.com/d/a/e/daeab7f5d64d3807059fc4096fe4ee360407b898.png)
![$\boxed{[DEF]\le\frac S4}$](//latex.artofproblemsolving.com/7/9/2/7922bf6d1739207f2739a9a93e4ac1869d173172.png)
. It is well-known or show easily that ![$\left\{\begin{array}{c}
S=[ABC]=2R^2\sin A\sin B\sin C\\\\
s_{\mathrm{orth}}=2R\sin A\sin B\sin C\end{array}\right|$](//latex.artofproblemsolving.com/0/2/4/0249ecb7dd7fb4e30fb4a0110bc3169cece5206c.png)
and the perimeter of 
is at least equally to the
perimeter
of the orthic triangle for . Thus, ![$\rho =\frac {[DEF]}{s_{DEF}}\le$](//latex.artofproblemsolving.com/3/b/e/3be22dcfb0f04c4840b1274742e6f0cfffbff5f2.png)
.
for which the projection 
of 
on 
belongs to the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
. Then 
.Proof. Apply the generalized Pythagoras' relation
. Otherwise. 
.Can show shorter with trigonometry (the Cosinus' theorem). Indeed,
.Remark. Using relation
can obtain Heron's formula. Indeed, 
![$\left[2ac+\left(a^2+c^2-b^2\right)\right]$](http://latex.artofproblemsolving.com/5/3/9/5398d45be7c21ab05a2b188797f95d8c004857ce.png)
![$\left[2ac-\left(a^2+c^2-b^2\right)\right]=$](http://latex.artofproblemsolving.com/b/f/d/bfdfc4fc848dbcc5988195755ed76fe41571f51a.png)
![$\left[(a+c)^2-b^2\right]$](http://latex.artofproblemsolving.com/e/d/f/edf1e55b53d83284d19bbabdf0b8835e1f45ea95.png)
![$\left[b^2-(a-c)^2\right]\implies$](http://latex.artofproblemsolving.com/5/a/0/5a0ddbd333ec541f42121e09b0bbc495b1319979.png)
.Remark. Let
![$[AL]$](http://latex.artofproblemsolving.com/7/3/7/737a9b49658cbe239832952bb575c8bbdf7dbb35.png)
be a diameter of the circumcircle 
. Thus, 
. in conclusion, 
. Hence 
.PP1 (grade 7 - modified enunciation). Let an
-right with 
. Denote 
so that 
,
and 
. Prove that 
and in this case find the ratios 
and 
. Proof 1. Show easily that
and 
. Apply the upper lemma: 
. Obtain analogously that 
.Method 1.1. I"ll use the property
. Hence 
. Using substitution 
, relation 
becomes 
Method 1.2. Denote
. Apply the Menelaus' theorem to the transversals 
in the triangles 
respectively:
.
.Therefore,
.Suppose that
. In this case 
. Apply the property 
. Apply the Menelaus' theorem to
and 
. Thus, 
.I"ll show that
. Indeed, apply the Menelaus' theorem to 
. Therefore, 
. See here.Proof 2 (analytic). Let
. Thus, 
. So 
Lemma 2. Let
and 
. Then there is the relation 
.Proof 3.. Let
. Thus, 
, 
are cyclic, i.e. 
and 
. Applylemma 2 and get
. I"ll use method 1.1 
.PP2 (grade 7). Let
with 
and 
and 
so that 
. Prove that 
.Proof.
belong to the 
-median 
.Thus, 
. Hence . This problem is very easy.PP3 (grade 8). Ascertain
and 
so that ![$[x] +\left\{\frac 1x\right\}=1.005\cdot n$](http://latex.artofproblemsolving.com/9/8/a/98a5c72fe761f445af22b5a25ef0d6591f6fc18a.png)
.Proof. Show easily that
. Hence 
, i.e. 
. Denote 
, where 
and 
. Thus, the equation becomes![$[x] +\left\{\frac 1x\right\}=\left(1+\frac 1{200}\right)\cdot n\iff$](http://latex.artofproblemsolving.com/1/1/7/1174fac7d1204b66f0ce59275aa8574fda4fff80.png)
![$[x] +\left\{\frac 1x\right\}=n+q+\frac r{200}$](http://latex.artofproblemsolving.com/b/f/5/bf50fb40eec6e67a0ce50f5807f9531a71859077.png)
. Hence ![$[x]=n+q$](http://latex.artofproblemsolving.com/8/c/a/8cadaca3761d3b8d94d27352827834251b79211b.png)
, i.e. ![$[x]=201q+r$](http://latex.artofproblemsolving.com/4/b/b/4bb540e825b667fb493a8ba8e5412ff726f8307a.png)
and 
. Thus, 
, i.e.
and 
. From ![$[x]\le x<[x]+1$](http://latex.artofproblemsolving.com/c/8/8/c88cb23bfa4209b9fd8de2b154a3c5402ea5d268.png)
obtain that 
.Since
and 
obtain that 
and 
and 
.PP4 (grade 9). Let an interior point
of for which denote 
. and the radius-length 
of the incircle for 
. Prove that ![$[DEF]\le \frac S4$](http://latex.artofproblemsolving.com/6/5/e/65ee17f34b37af3b87ba240cb12f33350a66b3b6.png)
and 
.Proof. Show easily that
, where 
![$[DEF]=\frac {2S}{(x+1)(y+1)(z+1)}\le$](http://latex.artofproblemsolving.com/d/a/e/daeab7f5d64d3807059fc4096fe4ee360407b898.png)
![$\boxed{[DEF]\le\frac S4}$](http://latex.artofproblemsolving.com/7/9/2/7922bf6d1739207f2739a9a93e4ac1869d173172.png)
. It is well-known or show easily that ![$\left\{\begin{array}{c}
S=[ABC]=2R^2\sin A\sin B\sin C\\\\
s_{\mathrm{orth}}=2R\sin A\sin B\sin C\end{array}\right|$](http://latex.artofproblemsolving.com/0/2/4/0249ecb7dd7fb4e30fb4a0110bc3169cece5206c.png)
and the perimeter of 
is at least equally to theperimeter
of the orthic triangle for . Thus, ![$\rho =\frac {[DEF]}{s_{DEF}}\le$](http://latex.artofproblemsolving.com/3/b/e/3be22dcfb0f04c4840b1274742e6f0cfffbff5f2.png)
.This post has been edited 85 times. Last edited by Virgil Nicula, Nov 15, 2015, 3:20 PM
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