Problem:
The vertices of △ABC are A=(0,0),B=(0,420), and C=(560,0). The six faces of a die are labeled with two A's, two B's, and two C's. Point P1​=(k,m) is chosen in the interior of △ABC, and points P2​,P3​,P4​,… are generated by rolling the die repeatedly and applying the rule: If the die shows label L, where L∈{A,B,C}, and Pn​ is the most recently obtained point, then Pn+1​ is the midpoint of Pn​L​. Given that P7​=(14,92), what is k+m?
Solution:
First note that, since P1​ is inside △ABC, all subsequent points Pk​ will also be inside the triangle. Furthermore, as will be shown below, once any subsequent Pk​ is given, then P1​ is uniquely determined. Suppose that Pk​=(xk​,yk​) is known. Since Pk​ is inside △ABC, we have
0<xk​<560,0<yk​<420,0<420xk​+560yk​<420⋅560
If A is rolled, then
(xk+1​,yk+1​)=Pk+1​=21​Pk​=(2xk​​,2yk​​),
so the range of possible positions of Pk+1​ is limited to the original triangle contracted by a factor of 21​ (region I in the diagram). Hence if A is rolled, then Pk+1​ is in the interior of region I, and we may conclude that
420xk+1​+560yk+1​<21​⋅420⋅560
Similarly, if B is rolled, then Pk+1​ is in the interior of region II, so yk+1​>210. If C is rolled, then Pk+1​ is in the interior of region III, so xk+1​>280. Thus, for k≥2,Pk​ must lie in one of the regions I,II,III, and its predecessor is uniquely determined. For example if Pk​=(xk​,yk​) lies in region II, then Pk​ must be the midpoint of BPk−1​​. It follows that Pk−1​=2Pk​−B=(2xk​,2yk​−420). We can now construct a "predecessor function" as follows: if k≥2 and Pk​=(xk​,yk​), then
Pk−1​=⎩⎪⎪⎨⎪⎪⎧​(2xk​,2yk​−420) if yk​>210(2xk​−560,2yk​) if xk​>280(2xk​,2yk​) if 420xk​+560yk​<21​420⋅560​
It is now easy to trace P7​=(14,92) back to P1​:
P7​=(14,92)⟹P6​=(28,184)⟹P5​=(56,368)⟹P4​=(112,316)⟹P3​=(224,212)⟹P2​=(448,4)⟹P1​=(336,8).​
We then see that k+m=336+8=344​.
The problems on this page are the property of the MAA's American Mathematics Competitions
