tag:blogger.com,1999:blog-6933544261975483399.post5643486010472633445..comments2022-09-27T03:11:10.165-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1350: Triangle with three Intersecting Circles, Incircle, Collinear PointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-86030337844985575312017-12-26T20:15:38.842-08:002017-12-26T20:15:38.842-08:00Zakrzew You got an excellent solution ! Peter tR...Zakrzew<br /><br />You got an excellent solution !<br /><br />Peter tRanPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-19559570744863919702017-12-26T12:36:48.775-08:002017-12-26T12:36:48.775-08:00All solutions posted here seems to be a bit to com...All solutions posted here seems to be a bit to complex.<br />Its easy to see A_4,B_4,C_4 all lie on power axis of incircle (i) and circum crircle of ABC (o)<br /> for example A_4 lies on power axis of i and A_1 (line A_2A_3)<br /> as well as on power axis of o and A_1 (line BC)<br /> So A_4 lies on power axis of o and i.<br /> Same for B_4 and C_4.Anonymoushttps://www.blogger.com/profile/14416907270784625951noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-31580686498459513472017-12-13T22:15:21.983-08:002017-12-13T22:15:21.983-08:00Referring to Problem 1350 and its figure. Let J be...Referring to Problem 1350 and its figure.<br />Let J be the center of the circle through B, C, and I<br />and R&#39; be the radius.<br />Do the the following results hold good ?<br />(i) R&#39; = R / 2 cos A<br />(ii)&lt;OJI = C - A<br />(iii) OI = R sin [(C - A)/2]/cos A <br /><br />Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-43237426374334904422017-12-13T20:30:59.983-08:002017-12-13T20:30:59.983-08:00Pravin: The above post is only a detailed version ...Pravin: The above post is only a detailed version of my earlier post. Thank you.Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63101400677333723252017-12-12T22:50:53.465-08:002017-12-12T22:50:53.465-08:00Problem 1350 Denote the points of contact of the i...Problem 1350<br />Denote the points of contact of the incircle with the sides BC, CA, AB as P, Q, R respectively. We first evaluate the ratio BA4:A4C in terms of a, b, c, and s (respectively the lengths of the sides BC, CA, AB and the semi perimeter of the triangle). Cyclically permuting we get the values of the similar ratios CB4:B4A and AC4:C4B. We next show that the product of these 3 ratios is -1, from which fact follows the collinearity of the points C4, A4, and B4 by the converse of Menelau’s Theorem.<br />First consider the in-circle I for which A4P is a tangent and A4 A3 A2 is a secant. By the Tangent – Secant Property we get A4P^2 = A4A3. A4A2 <br />Next consider the circle A1 for which both A4 A3 A2 and A4CB are secants. By the Secants Property we get A4A3 . A4A2 = A4C . A4 B. <br />Hence it follows A4C . A4B = A4B3^2. It is well known that CP = s - c<br />At this stage we let A4C = x . So A4P = A4C + CP = x + s -c<br /> So x(x + a) = (x + s - c)2, <br />Cancelling x^2 on either side we get xa = (s – c)^2 + 2(s-c)x,<br />x(a + 2c – 2s) = (s – c)^2, <br />x( c – b) = (s – c)^2,<br />x = (s – c)^2 / (c – b) ......(i)<br />Also 4(x + a)(c – b) = 4x(c – b) + 4a(c – b) <br />= 4(s – c)^2 + 4a(c – b) = (2s – 2c)^2 + 4ac – 4ab <br />= (a + b –c)^2 + 4ac – 4ab <br />= a^2 + b^2 + c^2 + 2ab -2bc – 2ca + 4ac – 4ab<br />= a^2 + b^2 + c^2 – 2ab – 2bc + 2ac = (c + a – b)^2<br />= (2s – 2b)^2= 4(s – b)^2, So <br />x + a = (s – b)^2 / (c – b) .......(ii)<br />Hence from (i) and (ii)<br />BA4 / A4C = (x + a)/x = (s – b)^2 /(s – c)^2 in magnitude.<br />Similarly by Cyclic Symmetry:<br />CB4 / B4A = (s - c)^2 / (s – a)^2 ,<br />AC4 / C4B = (s – a)^2 / (s – b)^2<br />With due regard to sign we obtain on multiplication:<br /><br />[BA4 / A4C] [CB4 / B4A] [AC4 / C4B] = -1<br />Hence by Menelau’s Theorem<br />C4, A4, and B4 are collinear.<br />Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-71046519129341961992017-12-12T02:38:04.171-08:002017-12-12T02:38:04.171-08:00Let T1,T2,T3 be the points of contact of the in-ci...Let T1,T2,T3 be the points of contact of the in-circle <br />with the sides BC,CA,AB resp.<br />Let A4T1=x,B4T2=y,C4T3=z <br />By Tangent-Secant property: <br />x(x+a)=A4C.A4B=A4A3.A4A2 referred to circle A1 while <br />A4A3.A4A2=A4T1^2 referred to in-circle.<br />Note A4T1=A4C+CT1=x+(s-c)<br />So x(x+a)=[x+(s-c)]^2,<br />xa=(s-c)^2 + 2x(s-c),<br />x(a+2c-2s)=(s-c)^2,<br />x(c-b)=(s-c)^2,<br />x=(s-c)^2/(c-b) and <br />x+a=(s-b)^2/(c-b)on further simplification.<br />Hence A4C/A4B=(s-c)^2/(s-b)^2 <br />By cyclic symmetry B4C/B4A=(s-a)^2/(s-c)^2 <br />and C4A/C4B=(s-b)^2/(s-a)^2<br />Product of the above 3 ratios being 1 (numerically), <br />by Menealau C4,A4,B4 are collinear.<br />Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47798138846251987012017-12-06T05:45:02.124-08:002017-12-06T05:45:02.124-08:00https://photos.app.goo.gl/Yubn9z3lzvSmV5w03 Denot...https://photos.app.goo.gl/Yubn9z3lzvSmV5w03<br /><br />Denote A’, B’, C’ as points of contact of incircle I to triangle ABC<br /> a, b,c and 2p as 3 sides and perimeter of triangle ABC<br />Denote k= B’C/B’A , s= C’A/C’B and q=A’C/A’B<br />Define point B5 on line AC such that B4B’=B4B5 ( see sketch)<br />Since B4 is on the radical line of circles I and B1 so <br />So B4A.B4C=B4B’^2=B4B5^2 =&gt; ( ACB’B5)= -1<br />And k= (B’C/B’A)=(B5C/B5A) =&gt; B’C/k=B’A/1=b/(1+k) =&gt; B’A= b/(1+k)<br />Similarly B5A=b/(1-k)<br />We have AB4= ½(AB’+AB5)= b/(1-k^2) and CB4=AB4- b= b.k^2/(1-k^2)<br />And B4C/B4A= k^2<br />Similarly we also have C4A/C4B= s^2 and A4B/A4C= 1/q^2<br />And (B4C/B4A) x(C4A/C4B)x(A4B/A4C)= k^2 . s^2/q^2<br />Replace k=(p-c)/(p-a) and s=(p-a)/(p-b) , q=(p-c)/(p-b) in above expression<br />We will have (B4C/B4A) x(C4A/C4B)x(A4B/A4C)= 1<br />And A4, B4 ,C4 are collinear per Menelaus’s theorem <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com