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| Line 1: |
Line 1: |
| | In an acute triangle <math>ABC</math>, let <math>M</math> be the midpoint of <math>\overline{BC}</math>. Let <math>P</math> be the foot of the perpendicular from <math>C</math> to <math>AM</math>. Suppose that the circumcircle of triangle <math>ABP</math> intersects line <math>BC</math> at two distinct points <math>B</math> and <math>Q</math>. Let <math>N</math> be the midpoint of <math>\overline{AQ}</math>. Prove that <math>NB=NC</math>. | | In an acute triangle <math>ABC</math>, let <math>M</math> be the midpoint of <math>\overline{BC}</math>. Let <math>P</math> be the foot of the perpendicular from <math>C</math> to <math>AM</math>. Suppose that the circumcircle of triangle <math>ABP</math> intersects line <math>BC</math> at two distinct points <math>B</math> and <math>Q</math>. Let <math>N</math> be the midpoint of <math>\overline{AQ}</math>. Prove that <math>NB=NC</math>. |
| | == Solution 1 == | | == Solution 1 == |
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| − | import graph; size(28.013771887739892cm);
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| − | real labelscalefactor = 0.5;
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| − | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
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| − | pen dotstyle = black;
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| − | real xmin = -1.278031073276777, xmax = 26.735740814463117, ymin = -9.456108920092317, ymax = 4.7809371214468275;
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| − | pen qqwuqq = rgb(0.,0.39215686274509803,0.);
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| − | draw((13.41604889400778,-6.856056531808553)--(13.294446243957523,-6.529198735935188)--(12.967588448084157,-6.650801385985445)--(13.089191098134414,-6.97765918185881)--cycle, linewidth(2.) + qqwuqq);
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| − | draw((9.638133559035012,2.2984960680144826)--(5.734005553668605,-4.81861693653532), linewidth(2.));
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| − | draw((5.734005553668605,-4.81861693653532)--(18.84453746580399,-4.836466959731463), linewidth(2.));
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| − | draw((18.84453746580399,-4.836466959731463)--(9.638133559035012,2.2984960680144826), linewidth(2.));
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| − | draw((9.638133559035012,2.2984960680144826)--(13.089191098134414,-6.97765918185881), linewidth(2.));
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| − | draw(circle((10.344923836854495,-2.718588274420166), 5.066624891969315), linewidth(2.));
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| − | draw((9.638133559035012,2.2984960680144826)--(14.950106647313914,-4.831164682073189), linewidth(2.));
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| − | draw((9.638133559035012,2.2984960680144826)--(9.628436372158689,-4.823919214193597), linewidth(2.));
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| − | draw((18.84453746580399,-4.836466959731463)--(13.089191098134414,-6.97765918185881), linewidth(2.));
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| − | draw((5.734005553668605,-4.81861693653532)--(13.089191098134414,-6.97765918185881), linewidth(2.));
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| − | draw((12.294120103174464,-1.2663343070293533)--(12.289271509736299,-4.827541948133391), linewidth(2.));
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| − | dot((9.638133559035012,2.2984960680144826),dotstyle);
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| − | label("<math>A</math>", (9.703893586940504,2.462896137778213), NE * labelscalefactor);
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| − | dot((5.734005553668605,-4.81861693653532),dotstyle);
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| − | label("<math>B</math>", (5.807611933540062,-4.655626882991359), NE * labelscalefactor);
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| − | dot((18.84453746580399,-4.836466959731463),dotstyle);
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| − | label("<math>C</math>", (18.910297493709486,-4.6720668899677325), NE * labelscalefactor);
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| − | dot((12.289271509736299,-4.827541948133391),linewidth(4.pt) + dotstyle);
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| − | label("<math>M</math>", (12.350734710136587,-4.688506896944105), NE * labelscalefactor);
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| − | dot((13.089191098134414,-6.97765918185881),linewidth(4.pt) + dotstyle);
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| − | label("<math>P</math>", (13.156295051978871,-6.842147810848988), NE * labelscalefactor);
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| − | dot((14.950106647313914,-4.831164682073189),linewidth(4.pt) + dotstyle);
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| − | label("<math>Q</math>", (15.01401584030904,-4.7049469039204785), NE * labelscalefactor);
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| − | dot((12.294120103174464,-1.2663343070293533),linewidth(4.pt) + dotstyle);
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| − | label("<math>N</math>", (12.36717471711296,-1.1374653900475058), NE * labelscalefactor);
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| − | dot((9.628436372158689,-4.823919214193597),linewidth(4.pt) + dotstyle);
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| − | label("<math>X</math>", (9.687453579964131,-4.688506896944105), NE * labelscalefactor);
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| − | label("<math>\alpha = 90^\\circ</math>", (13.156295051978871,-6.792827789919868), NE * labelscalefactor,qqwuqq);
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| − | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
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| − |
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| | Let <math>X</math> be the foot from <math>A</math> to <math>\overline{BC}</math>. By definition, <math>\angle AXM = \angle MPC = 90^{\circ}</math>. Thus, <math>\triangle AXM \sim \triangle MPC</math>, and <math>\triangle BMP \sim \triangle AMQ</math>. | | Let <math>X</math> be the foot from <math>A</math> to <math>\overline{BC}</math>. By definition, <math>\angle AXM = \angle MPC = 90^{\circ}</math>. Thus, <math>\triangle AXM \sim \triangle MPC</math>, and <math>\triangle BMP \sim \triangle AMQ</math>. |
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In an acute triangle 
, let 
be the midpoint of 
. Let 
be the foot of the perpendicular from 
to 
. Suppose that the circumcircle of triangle 
intersects line 
at two distinct points 
and 
. Let 
be the midpoint of 
. Prove that 
.
Solution 1
Let 
be the foot from 
to . By definition, 
. Thus, 
, and 
.
From this, we have 
, as 
. Thus, is also the midpoint of 
.
Now, 
iff lies on the perpendicular bisector of . As lies on the perpendicular bisector of 
, which is also the perpendicular bisector of (as is also the midpoint of ), we are done.
~ Martin2001, ApraTrip
, let 
be the midpoint of 
. Let 
be the foot of the perpendicular from 
to 
. Suppose that the circumcircle of triangle 
intersects line 
at two distinct points 
and 
. Let 
be the midpoint of 
. Prove that 
.
be the foot from 
to 
. Thus, 
, and 
.
, as 
. Thus, 
.
iff 
, which is also the perpendicular bisector of 
