Geometry Problem 1542. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

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ReplyDeleteSee sketch for location of points P, Q,R and angles u, v and radius r

Note that triangles AFM, NEM and CDN are similar ( case AA)

And OM and ON are angle bisectors of angles POR and QOR= u+v= 45

We have ME/MF= MN/MA= 6/16 = ME/(2r- ME) => ME= 6/11. r

And PM= 5/11.r

tan(u+v)=( tan(u)+tan(v))/(1- tan(u).tanv))= 1,,,,, (1)

Replace tan(u)= PM/r= 5/11 in (1)

We get tan(v)=⅜= NQ/r => NQ= ⅜.r and NE= ⅝ . r and ND= 11/8 .r

We have NC/NM= ND/NE=(1.375 r)/(.625 r)= 11/5

So NC= 66/5

Let NC=x,

ReplyDeleteConsider triangle AOC,

Angle AOC=135 degrees,

Angle AOM = Angle MON = Angle NOC= 45 degrees,

ON is bisector of Angle MOC hence

OM/OC=MN/NC=6/x,

And OA is external bisector of Angle MOC, hence OM/OC=AM/AC=16/(x+22)

We get 6/x = 16/(x+22),

Simplifying it we get x=6×22/10=13.2 units.

NC = 11.2

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