Proposed Problem
Click the figure below to see the complete problem 435 about Triangle, Incircle, Inradius, Congruent circles, Tangent, Measurement, Similarity.
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Complete Problem 435
Level: High School, SAT Prep, College geometry
Sunday, April 18, 2010
Problem 435: Triangle, Incircle, Inradius, Congruent circles, Tangent, Measurement
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Let's denote centres of extreme left & extreme right inner circles as P and Q. PQ is clearly parallel to AC and AP & CQ are internal bisectors of Tr, ABC and will therefore meet in I, the in-centre. Now the altitude from I to BC is r while the distance between PQ and AC is x. Hence, altitude of Tr. IPC from I to PQ = r-x. Triangles IPQ and IAC are similar and thus PQ/AC = ratio of their altitudes =(r-x)/r. But PQ = 8x and AC=b. Thus, 8x/b =(r-x)/r or x = br/(8r+b)
ReplyDeleteAjit
Line 4 Typo -- Should read Tr. IPQ
Deletesolution of joe is nice
ReplyDeleteanother, just for working with ratios
M, G, N tg points of left, big, right circles
▲APM ~ ▲AIG, ▲CQN ~ ▲CIG
x/r = AM/AG, x/r = CN/CG
AM/AG = CN/CG => AM/CN = AG/GC => (AM+CN)/CN=(AG+GC)/GC
=>(b-8x)/CN = b/GC => (b-8x)/b = CN/GC => (b-8x)/b=x/r
=>
x = br/(b+8r)
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