Problem:
The sides of △ABC have lengths 6,8 and 10. A circle with center P and radius 1 rolls around the inside of △ABC, always remaining tangent to at least one side of the triangle. When P first returns to its original position, through what distance has P traveled?
Answer Choices:
A. 10
B. 12
C. 14
D. 15
E. 17
Solution:
When the circle is closest to A with its center P at A′, let its points of tangency to AB and AˉCˉ be D and E, respectively. The path parallel to AB is shorter than AB by AD plus the length of a similar segment at the other end. Now AD=AE=cot2A. Similar reasoning at the other vertices shows that the length of the path of P is
AB+BC+CA−2cot2A−2cot2B−2cot2C.
Since cot2A=sinA1+cosA=531+54=3, and similarly cot2B=1 and cot2C=2, the length of the path is 8+6+10−6−2−4=12.
OR
The locus of P is a triangle A′B′C′ similar to triangle ABC. Calculating as above, we have A′B′=AB−cot2A−cot2B=8−3−1=4=21AB, so the linear dimensions of △A′B′C′ are half those of △ABC, and its perimeter is 28+6+10=12.
OR
The locus, △DEF, of the center of the rolling circle is similar to △ABC, so we label its sides 3x,4x and 5x, for some x>0. The area of △ABC is (AB)(BC)/2=24. Partition △ABC into three trapezoids of altitude 1 and △DEF, and compute the area of △ABC in terms of x:
[ABC]=[DABE]+[EBCF]+[FCAD]+[DEF]=21(1)(4x+8)+21(1)(3x+6)+21(1)(5x+10)+6x2=6x2+6x+12.
Solve 6x2+6x+12=24 for the positive root, x=1, to find that the perimeter of △DEF is 3x+4x+5x=12x=12.
Challenge. Prove that for any triangle and for any circle that rolls around inside the triangle, the perimeter of the triangle, which is the locus of the center of the circle, is the perimeter of the original triangle diminished by the perimeter of the similar triangle, which circumscribes the circle.
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