1982 AHSME Problems/Problem 24

Solution 1

[asy] defaultpen(fontsize(10)); real r=sqrt(22); pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J); dot(A^^B^^C^^D^^E^^F^^G^^H^^J); draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7)); label("$A$", A, N); label("$B$", B, dir(210)); label("$C$", C, dir(330)); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, dir(170)); label("$G$", G, dir(250)); label("$H$", H, SE); label("$J$", J, dir(0)); label("2", A--G, dir(30)); label("13", F--G, dir(180+30)); label("1", F--C, dir(30)); label("7", H--J, dir(-30));[/asy]

Let $AH=y, BD=a, DE=x,$ and $EC=b.$ Because $AG = 2, GF = 13, HJ = 7$ and $FC = 1$ and they sum to 16, the length of each side of the equilateral triangle will be 16. By Power of a Point on A, we have $y(y + 7) = 2(2 + 13),$ or expanding and factoring, $0 =y^2 + 7y - 30 = (y - 3)(y + 10).$ Therefore, y must equal 3 as this triangle must have positive side lengths.

Hence, $BJ=16-(3+7)=6.$

Power of a point on C yields $b(b + x) = 1(1 + 13) = 14,$ and power of a point on B yields a(a + x) = 6(6 + 7) = 78. Also, we know that because a, b, and x are non-intersecting segments of the side of the triangle, a + b + x = 16.

So, we have a system of equations:

\[a^2 + ax = 78,\] \[b^2 + bx = 14,\] $a + b + x =16.$

If we subtract the second equation from the first, it appears to be a difference of squares! And we like differences of squares, because we can factor! Subtracting yields $a^2 - b^2 + (a - b)x = (a-b)(a+b)(a-b)x=(a - b)(a + b + x).$ Because we know that $a+b+x=16,$ inputting this we have that $(a - b)16 = 78 - 14 = 64.$ Therefore $a-b=4.$ Then adding this to the third equation, 2a + x = 20, so $a = 10 - (x/2).$ Substituting this into #1, we can now solve for x, and we have a difference of squares, or $100-x^2/4=78.$ This yields $x^2/4=22,$ so $x/2=\sqrt {22}$ and $x= \left(A\right)2\sqrt{22}!$

Further clarifications: If you are unsure why we didn't solve for a and b, it is because it is more efficient to use clever manipulations like the ones we used in this problem. We were looking for sums of values we already know, so we indirectly implied dividing $a^2 - b^2 + (a - b)x$ by $a+b+x$ to get $a-b,$ conveniently, then everything fell into place with substitution.

~*Solution by ab2024