Točno
19. studenoga 2017. 13:13 (6 godine, 8 mjeseci)
Let
be a triangle with circumcentre
. The points
and
are interior points of the sides
and
respectively. Let
and
be the midpoints of the segments
and
. respectively, and let
be the circle passing through
and
. Suppose that the line
is tangent to the circle
. Prove that
Proposed by Sergei Berlov, Russia
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
![O](/media/m/9/6/0/9601b72f603fa5d15addab9937462949.png)
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
![Q](/media/m/4/5/c/45ce8d14aa1eb54f755fd8e332280abd.png)
![CA](/media/m/a/a/e/aaec86bc003cfdb64d54116a4cabd387.png)
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
![K,L](/media/m/c/f/8/cf8b7b8c56970a06671ff82ddb7f6450.png)
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
![BP,CQ](/media/m/6/2/f/62f9112941d60283b09cdc1c5b2874ad.png)
![PQ](/media/m/f/2/f/f2f65ec376294df7eca22d2c1a189747.png)
![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
![K,L](/media/m/c/f/8/cf8b7b8c56970a06671ff82ddb7f6450.png)
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
![PQ](/media/m/f/2/f/f2f65ec376294df7eca22d2c1a189747.png)
![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
![OP = OQ.](/media/m/c/0/1/c0135f985d771c12d363a4df1deeb689.png)
Proposed by Sergei Berlov, Russia
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Uočimo prvo da nam polovišta daju dosta informacija o konfiguraciji, imamo
odnosno
odakle slijedi
iz uvjeta tangentnosti
odnosno imamo slijedeće zaključke ![\angle AQM = \angle KLM \quad \textrm{i analogno} \quad \angle APM= \angle LKM](/media/m/4/2/b/42b4f8e9d6df1445f22da7d5b0b37179.png)
Što implicira![\triangle AQP \sim \triangle KLM](/media/m/6/a/e/6ae8f9933eb445ab581a02f5b33f0cb2.png)
Pa povlačimo slijedeće odnose![\frac{AQ}{ML}=\frac{AP}{MK} , \quad \quad \frac{PC}{ML}=\frac{QB}{MK}](/media/m/4/4/1/4410168018de8bb443e05284d05a00af.png)
Prvi odnos slijedi iz sličnosti danih trokuta, dok drugi slijedi iz prvobitnog zaključka o paralelnosti odnosno da je
srednjica trokuta
. Kombiniranjem dobivenog imamo
dakle
i
imaju jednaku potenciju na opisanu kružnicu
.
![R^2-OQ^2=R^2-OP^2](/media/m/6/1/2/6129c027e25625f9b9400dd616c9bf7a.png)
Odnosno
![KM||BQ](/media/m/5/f/0/5f05c97e814d4b870667d05e7f159ef8.png)
![KM||AB](/media/m/0/5/7/057683a9fc9a7c32e20d02f58b7c8e0d.png)
![\angle AQM = \angle QMK](/media/m/b/2/e/b2ef74f23cae20fa151582ca8afde110.png)
![\angle QMK=\angle KLM](/media/m/f/f/8/ff80ce202b6b167aa11b1ac7b5c17efa.png)
![\angle AQM = \angle KLM \quad \textrm{i analogno} \quad \angle APM= \angle LKM](/media/m/4/2/b/42b4f8e9d6df1445f22da7d5b0b37179.png)
Što implicira
![\triangle AQP \sim \triangle KLM](/media/m/6/a/e/6ae8f9933eb445ab581a02f5b33f0cb2.png)
Pa povlačimo slijedeće odnose
![\frac{AQ}{ML}=\frac{AP}{MK} , \quad \quad \frac{PC}{ML}=\frac{QB}{MK}](/media/m/4/4/1/4410168018de8bb443e05284d05a00af.png)
Prvi odnos slijedi iz sličnosti danih trokuta, dok drugi slijedi iz prvobitnog zaključka o paralelnosti odnosno da je
![MK=\frac{1}{2}QB](/media/m/5/5/8/558ac1a5f7788c49d915edd13ee39e5b.png)
![\triangle QBP](/media/m/b/e/9/be912c4b575fcbd4714cd3ed295c0c18.png)
![QA\cdot QB =PA \cdot PC](/media/m/6/0/3/60380063cce07f79c61d9bfedf1aa7df.png)
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
![Q](/media/m/4/5/c/45ce8d14aa1eb54f755fd8e332280abd.png)
![\triangle ABC](/media/m/1/f/3/1f3c3c0f3e134a169655f9511ba6ea82.png)
![Pow_{\Omega}(Q)=Pow_{\Omega}(P)](/media/m/7/8/2/78285e3ec8aa6f680da1ee6780140f24.png)
![R^2-OQ^2=R^2-OP^2](/media/m/6/1/2/6129c027e25625f9b9400dd616c9bf7a.png)
Odnosno
![OP=OQ](/media/m/9/f/b/9fbb80b5a64e7f7b659f30b5ee5febea.png)