Točno
9. ožujka 2018. 00:41 (6 godine, 4 mjeseci)
Let be a triangle with circumcircle
and incentre
. Let the line passing through
and perpendicular to
intersect the segment
and the arc
(not containing
) of
at points
and
, respectively. Let the line passing through
and parallel to
intersect
at
, and let the line passing through
and parallel to
intersect
at
. Let
and
be the midpoints of
and
, respectively. Prove that if the points
,
, and
are collinear, then the points
,
, and
are also collinear.
(U.S.A.)
Upozorenje: Ovaj zadatak još niste riješili!
Kliknite ovdje kako biste prikazali rješenje.
Kliknite ovdje kako biste prikazali rješenje.
Iskoristimo prvo paralelnosti koje su nam dane u zadatku tj. ![\angle YVI = \angle XUI = \angle (AI, BC) = 180 - \frac{\alpha}{2} - \gamma = \frac{\beta}{2} = \angle ABI = \angle YBI](/media/m/f/5/9/f59d5697cacc7fdb41deff712bbd9c64.png)
Dakle četverokut
je tetivan, nadalje motivirani ovom tetivnošću definiramo presjek pravaca
i
kao točka
. Sada kako je
i
imamo da je
također tetivan. Kratkim angle chaseom ovog kuta dobijamo
odakle slijedi da je
jednakokračan. U ovom trenutku se jos nije razvila neka posebna ideja za rijesiti zadatak pa nastavljamo sa iskorištavanjem uvjeta, tj uvjeta kolinearnosti točaka
i
. Uz paralelnost dobijamo:
Ili možda u korisnijoj formi
Sada u biti uviđamo neku simetriju sa već definiranom okomicom iz
na
.
Problem je kako u skicu na lijep način ubaciti polovišta jer ne izgleda kao da možemo
i
nekako pametno povezati.
Sljedeći korak je svakako plodan: definiranje točke
simetričnoj točki
preko
, tu je
polovište dužine
odakle je
. Sada smo već nekako zgrabili točke
i
u neki dio skice, i odmah se možemo lakše igrati... ostaje naravno dokazati
što bi riješilo zadatak.
(Arseniy Akopyan EnGeoFigures) prvi dan. Dodajmo polovište
luka
sa točkom
. Lemma glasi da je ![\angle IZB = \angle IMA](/media/m/a/b/8/ab8223b07a3d8ed2fdbd3c71867a1daf.png)
Dokaz lemme ostavljen čitatelju.
Uočimo sličnost trokutova
i
. Trivijalno je
¸Ne toliko teško dolazimo do jednakosti
odakle zbog sličnosti slijedi
dakle
raspolavlja luk
sa
u točki
. Neka je još
sjecište
i
. Imamo
pa
tetivan time je uz
i prije navedeno
dakle
odnosno
kolinearne
![\angle YVI = \angle XUI = \angle (AI, BC) = 180 - \frac{\alpha}{2} - \gamma = \frac{\beta}{2} = \angle ABI = \angle YBI](/media/m/f/5/9/f59d5697cacc7fdb41deff712bbd9c64.png)
Dakle četverokut
![YBVI](/media/m/2/2/a/22a7e94a11ec54450b16a398ebe7d222.png)
![UX](/media/m/7/5/b/75b39f8418fd82a2f47179dca2917a3d.png)
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
![N](/media/m/f/1/9/f19700f291b1f2255b011c11d686a4cd.png)
![\angle XUI = \beta/2](/media/m/b/2/6/b26c919e09ab5c8f0aab12674bda50be.png)
![\angle NUI= \beta/2](/media/m/3/0/3/3031222787b55d439139d0949a2c9ca6.png)
![NBUI](/media/m/6/4/e/64eb97955b227f841b9c90ebfafbcb02.png)
![\angle UNI = \angle UBI = \angle CBI = \beta/2 = \angle XUI = \angle NUI](/media/m/b/b/a/bbad732598f9d3b47d8149d0c90478c5.png)
![\triangle NUI](/media/m/9/e/7/9e7ecca5e7416b83d36d4530f3d99231.png)
![I, X](/media/m/f/c/9/fc9f3739750c01d0ba8ad6da7c728af6.png)
![Y](/media/m/3/b/c/3bc24c5af9ce86a9a691643555fc3fd6.png)
![\frac{XU}{YV}=\frac{IX}{IY}=\frac{AN}{AY}=\frac{NX}{YV} \Rightarrow](/media/m/a/a/6/aa633a140e52610a321d490a5c3a8374.png)
![XU=NX \Rightarrow IY \perp NU](/media/m/b/d/6/bd6c0c26800ef5703dbc29fbf9160150.png)
![YI \perp AI](/media/m/9/c/c/9cc35ac11b5ddfedd283830aa80ad2a2.png)
![I](/media/m/3/8/6/38689d6affa9ba35368ca4d3d76ea147.png)
![CI](/media/m/b/6/5/b656d89737b7825cdbf637e863735c23.png)
Problem je kako u skicu na lijep način ubaciti polovišta jer ne izgleda kao da možemo
![Z](/media/m/7/9/4/794ff2bd637e30ea27e50e57eecd0b76.png)
![W](/media/m/2/c/4/2c4aa0f61279d74f16a59bcde17578ef.png)
Sljedeći korak je svakako plodan: definiranje točke
![X'](/media/m/2/4/d/24de7e028e3b08c56dc4493496de8ffb.png)
![X](/media/m/9/2/8/92802f174fc4967315c2d8002c426164.png)
![I](/media/m/3/8/6/38689d6affa9ba35368ca4d3d76ea147.png)
![I](/media/m/3/8/6/38689d6affa9ba35368ca4d3d76ea147.png)
![XX'](/media/m/2/e/6/2e6482a981f03697f7d0308311df918a.png)
![IW || AX'](/media/m/6/7/f/67fe5daaa97a6d47da46d6a0fb10fb5b.png)
![W](/media/m/2/c/4/2c4aa0f61279d74f16a59bcde17578ef.png)
![I](/media/m/3/8/6/38689d6affa9ba35368ca4d3d76ea147.png)
![IZ || AX'](/media/m/f/4/9/f4931b2d00b511c9dd3968b395bbd7dc.png)
![Lemma:](/media/m/a/f/f/affec61aad5eb2770125833866b85e48.png)
![HMO- 2017](/media/m/0/b/0/0b0f00807f0798fba870a6476f2519b7.png)
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
![\angle IZB = \angle IMA](/media/m/a/b/8/ab8223b07a3d8ed2fdbd3c71867a1daf.png)
Dokaz lemme ostavljen čitatelju.
Uočimo sličnost trokutova
![\triangle AYX](/media/m/4/8/8/488e00d1279ecc4da9e28e922e661861.png)
![\triangle CUV](/media/m/5/8/6/5868b8dcad13b24bb22f2b4bc4484e62.png)
![\angle YAX = \angle UCV](/media/m/1/e/e/1ee9b11b09a37a6d75d46bbf4ad3eb31.png)
![\angle CUV = \angle AXY](/media/m/1/2/3/123c7c168892c07b0b0e873fc38165e6.png)
![\angle CVI = \angle CVU = \angle AYX = \angle BVI](/media/m/9/9/2/992ae7c20f3da7d9ec1aeb088f3bdf2c.png)
![VI](/media/m/f/9/8/f989d4da5dbc4358200611e10ffcad86.png)
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
![AX'](/media/m/e/4/c/e4c1a441d0d6c28bf4433892710dcdb6.png)
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
![\angle AX'I= \angle AIX' = \angle IUS](/media/m/b/5/1/b5136ef36c7abd553b510ba12df7d85e.png)
![IUSX'](/media/m/0/b/3/0b31e48107c2cb6de1ec58d1074a5ba7.png)
![lemmu](/media/m/1/0/e/10ef8021ca2d2a99b4f2fc3983c0c472.png)
![\angle IZB = \angle IMA = \angle VMA = \angle VIY = \angle ASB](/media/m/d/c/0/dc05f519ba73edc4d12834a753972a6e.png)
![IZ || AS || WI](/media/m/c/a/4/ca49eb90edeb25259193680ec91926fd.png)
![W,I,Z](/media/m/d/8/6/d8663003eadb8effe7fe66439c3969dd.png)