Circle Cutting an Equilateral Triangle

The side length of the equilateral triangle is 3+39+6=48. The power of A with respect to $\Gamma$ is $AD\cdot AE=3(3+39)=136$ . The power of B with respect to $\Gamma$ is $BD\cdot BE=(39+6)6=270$ . Let $BF=x$ , then by the power of a point theorem, $BF\cdot BG=270$ , so $x(x+21)=270$ . The only positive solution is $x=9$ . Since the side length is 48, $CG=48-21-9=18$ . Therefore, the power of C with respect to $\Gamma$ is $CG\cdot CF=18(18+21)=702$ .

Let $AI=a,\ CH=b$ , then by power of a point theorem $a(48-b)=136\qquad b(48-a)=702$ Subtract those two equations and divide by 48 to get $b-a=12$ . Substitute this into the first equation to get a quadratic in $a$ . The quadratic equation has solutions $a=18\pm 3\sqrt{22}$ , so $b=30\pm 3\sqrt{22}$ . Since $a+b<48$ , the solution with the negative sign is taken, giving $HI=48-a-b=6\sqrt{22}$ . Therefore $|HI|^2=792$ .

By Power of Point with respect to point $B$ , we have that $BE \cdot BD = BF \cdot BG$ , so $6\cdot45 = BF(BF+21)$ . It is easy to verify that the only positive solution is $BF = 9$ . Because $\triangle ABC$ is equilateral, we have that $AB = 3+39+6 = BC = 9 + 21 + GC$ , so the side length of $\triangle ABC$ is $48$ and $GC = 18$ .

Let $AI = a$ , $HI = b$ , and $CH = c$ . By Power of Point with respect to points $A$ and $C$ , we obtain: $a(a+b) = 3(3+39) = 3(42) = 126$ and $c(b+c) = 18(18+21) = 18(39) = 702$ . Note that $a+b+c = 48$ , so we can rewrite the two equations as: $a(48-c) = 126$ and $c(48-a) = 702$ .

Expanding both, we obtain $48a - ac = 126$ and $48c - ac = 702$ . Subtracting the two equations and dividing by $48$ , we obtain $c-a = 12$ , and so $c = a + 12$ and $a = c - 12$ .

Plugging in $c = a + 12$ into $a(48-c) = 126$ gives $a(36-a) = 126$ , or $a^2 - 36a + 126 = 0$ . By the quadratic formula, $a = \frac{36 \pm\sqrt{792}}{2}$ .

Plugging in $a = c - 12$ into $c(48-a) = 702$ gives $c(60-c) = 702$ , or $c^2 - 60c + 702 = 0$ . By the quadratic formula, $c = \frac{60 \pm \sqrt{792}}{2}$ .

Now consider these $\pm$ 's. It is easy to verify that both must be negative because $a+c < 48$ . Therefore, $a = \frac{36 - \sqrt{792}}{2}$ and $c = \frac{60 - \sqrt{792}}{2}$ . Then $b = 48 - a - b = \sqrt{792}$ , so the desired answer is $HI^2 = b^2 = 792$ .

Note $AB = AD + DE + EB = 3+39+6 = 48$ , which is the side length of equilateral $\triangle ABC$ . By the secant theorem, $BF \cdot BG = BE \cdot BD = 6(6+39),$ so $BF(BF + 21) - 270 = (BF + 30)(BF - 9) = 0$ . Hence $BF = 9$ and $CG = BC - BF - FG = 48 - 9 - 21 = 18$ . We apply the secant theorem again to obtain the system $\begin{aligned} CH(CH + HI) &= CG \cdot CF = 18(39), \\ AI(AI + HI) &= AD \cdot AE = 3(42), \\ CH + HI + IA &= 48. \end{aligned}$ This gives $CH (48 - IA) = 18(39)$ and $IA (48 - CH) = 3(42)$ , and subtracting the latter from the former yields $CH - IA = 12$ . Hence $48 = (IA + 12) + HI + IA$ , or $IA = 18 - \frac{HI}{2}$ . Finally, we obtain $\begin{aligned} 3(42) &= IA(HI + IA) = \left(18 - \frac{HI}{2}\right)\left(18 + \frac{HI}{2}\right) \\ &= 18^2 - \frac{HI^2}{4}, \end{aligned}$ or $|HI|^2 = 4 \cdot 3^2 (6^2 - 14) = \boxed{792}.$

We're going to use something called the Intersecting Secants Theorem (also known as the secant-secant theorem) which tells us that when two secant lines intersect each other outside a circle, the products of their segments are equal.

Now notice that $DB$ and $GB$ are two secants of the circle $\Gamma$ and they intersect at a point outside $\Gamma$ , $B$ .

So the intersecting secants theorem tells us:

$BD\cdot BE= BG\cdot BF$

$\Rightarrow (BE+BD)\cdot BE=(BF+FG)\cdot BF$

Plugging in all the values gives us:

$BF^2+21BF-270=0$ .

Solving this we get $BF=9$ .

Again notice that the sides of the equilateral triangle are equal to $48$ [because $AB=AD+DE+EB=3+39+6=48$ ].

So, $CG=48-FG-BC=48-21-9=18$ .

Again, $FC$ and $IC$ are two secants of the circle $\Gamma$ and they intersect at $C$ .

So, $CF\cdot CG= CI\cdot CH$

$\Rightarrow (21+18)\cdot 18 = CI\cdot CH$

$\Rightarrow 702 = CI\cdot CH$ $\cdots (1)$ .

We have another set of secants. Take $EA$ and $HA$ .

So, $AE\cdot AD= AH\cdot AI$

$\Rightarrow (3+39)\cdot 3= (48-CH)(48-CI)$

$\Rightarrow 126=48^2-48(CI+CH)+CI\cdot CH$

From $(1)$ , $CI\cdot CH=702$ .

So making this substitution, we get:

$126=2304-48(CI+CH)+702$

$\Rightarrow (CI+CH)=\frac{2304+702-129}{48}=60$ .

Squaring both sides gives us:

$(CI+CH)^2=60^2=3600$

$\Rightarrow (CI-CH)^2+4\cdot CI\cdot CH=3600$ .

Again recall $(1)$ which allows us to substitute $CI\cdot CH$ with $702$ . Also notice that $CI-CH=HI$ . So we're almost done!

Plugging in all the values gives us:

$HI^2=3600-4\cdot 702=\boxed{792}$ .

From power of a point on $B$ , we have $BE\cdot BD=BF\cdot BG$ , hence $6\cdot45=BF(BF+21)\Rightarrow BF=9$ . $48=AB=BC=9+21+GC$ , so $GC=18$ . Using power of a point on $C$ , we obtain $CG\cdot CF=CH\cdot CI\Rightarrow 18\cdot39=CH(CH+HI)$ . Similarly, we have $3\cdot42=AI(AI+HI)$ . Letting $CH=a, HI=b$ and $IA=c$ , we have $18\cdot39=a(a+b), 3\cdot42=c(c+b)$ and $a+b+c=48$ . Because $a+b=48-c$ and $b+c=48-a$ , we can substitute to get $702=a(48-c)$ and $126=c(48-a)$ . Subtracting these yields $12=a-c\Rightarrow a=12+c$ . Plugging this back into $126=c(48-a)$ gives us $126=c(36-c)\Rightarrow c=18-3\sqrt{22}$ , therefore $a=30-3\sqrt{22}$ . $HI^2=(48-(18-3\sqrt{22})-(30-3\sqrt{22}))^2=(6\sqrt{22})^2=\boxed{792}$ .

Let BF=x. By power of a point, (BE)(BD)=(BF)(BG). (6)(45)=x(x+21). After simplification and factorization, x=-30, 9. We discard -30 to find BF=9. The triangle's side length is 48, so 9+21+GC=48. GC=18. Let AI=a, HI=b, and CH=c. a+b+c=48. By power of a point, (3)(42)=a(a+b). Also, (18)(39)=c(b+c). Three variables, three equations: solve for b (and a/c along the way). Substituting 48-c for a+b and 48-a for b+c makes the process easier. After simplification, a=18-sqrt(198), c=30-sqrt(198). Therefore, b=sqrt(798). b^2=798, and we are done.

1st power of point B w.r.t. Circle is BE(BE+ED) OR also BF(BF+FG) . Hence solving this eqn. we get BF=9,as the tri. Is eqlateral GC=18 and

AI+HI+CH=48

let AI=x, HI=y,CH=z

applying eqn. Of power of point w.r.t. A and then C we get 2 eqn.

x^2 +xy-126=0.......1

z^2+zy-702=0..........2

subtracting these 2 eqn. And solving we get

z-x=12 .........3

we have z+x=48-y...........4

eq3 +eq4 we get

z = 30 - y/2

putting these value in eq. 2 we obtain-

y^2 =|HI|^2 =792

We use power of a Point from $BE$ . Since $BE*BD = BF * BG, 6*45 = BF * (BF+21)$ , so $BF = 9$ . Since $ABC$ is equilateral, $AB = 48$ , so $BC = AC = 48$ , which means that $GC = 18$ . By power of a point from A, we get that $3 * 42 = AI * (48-HC)$ , and power of a point from $C$ gives $18*39 = CH * (48 - AI)$ , so solving we have that $HI = \sqrt{792}$ so our answer is 792.

The secant secant theorem for the chords ED, GF and the outside point B, says that: |EB|⋅|DB|=|GB|⋅|FB|. |GB|=|FB|+|FG| => |EB|⋅|DB|=|FB|⋅(|FB|+|FG|) => |FB|⋅(|FB|+21)=6⋅(6+39)=270 => |FB|=9. ABC is an equilateral triangle => |AB|=|BC|=|CA|=|EB|+|DE|+|AD|=6+39+3=48=|FB|+|FG|+|GC| => |GC|=18. The secant secant theorem for the chords FG, HI and the outside point C: |GC|⋅|FC|=|HC|⋅|IC| =>|HC|⋅(|HC|+|HI|)=18⋅(18+21)=702, but |HC|+|HI|+|IA|=|AC|=48 => |HC|+|HI|=48-|IA|. Hence, |HC|⋅(48-|IA|)=702. (1) Applying the secant secant theorem again - this time for the chords DE, IH and the outside point A - we get that |DA|⋅|EA|=|IA|⋅|HA| => 3⋅(3+39)=|IA|⋅(|IA|+|HI|), but |IA|+|HI|=48-|HC| => |AI|⋅(48-|HC|)=126. (2) Substracting the relations (1) and (2), we obtain that 48⋅(|HC|-|IA|)=702-126=576 => |HC|-|IA|=12. Replacing |HC| with 12+|IA| in the second relation, we get that: |IA|⋅(48-12-|IA|)=126 => |IA|^2-36⋅|IA|+126=0 => |IA|=18-3√22. |HI|=48-|IA|-|HC|=48-|IA|-12-|IA|=36-2⋅|IA|=36-(36-6√22)=6√22. Hence, |HI|^2=36⋅22=792.

Using Secant Theoram

$\begin{aligned}(39+6)\cdot 6 &= BF \cdot (BF + 21)\\ \Rightarrow BF^2 + 21BF -270 &= 0\\ \text{And, BF is a length, hence } &\text{positive }\\ BF = 9~and~CG &= 18\\\\ \text{Now, }y(x + y ) &= 18 \times 39\\ \text{And, } (48-y)\times\left[48 - (x+y)\right] &= 42\times 3\\ \text{Solving.... }x&= \pm 6\sqrt{22}\\ \text{And, since the distance is positive, }&|HI| = 6\sqrt{22}\\\\ \boxed{|HI|^2 = 792}\end{aligned}$

Is it possible to determine the radius of this circle? I got a frightening number such as (7/19)(8479/3)^(1/2) or R ~=19.59. Can anyone please confirm this? If it's wrong what would be correct radius?

AB= 48 triangle BDF and BGE are similar. BE : BG = BF: BD and BF \times BG = BE \times BD BF \times (BF+21) = 270

=> BF=9 [ BF= -30 rejected]

CH \times CI = CG \times CF AI \times AH = AD \times AE note :- let CH = x , HI = y , IA = z

x \times (x+y) = 702 - eq 1. z \times (z+y) = 124 - eq 2. x+y+z = 48 - eq 3.

from eq 3. z= 48-x-y

and from eq 1. y = \frac { (702 - x^2)}{x}

putting value of z and y in eq .2

(48-x-y) \times ( 48-x-y+y) = 124 also, putting value of y in upper eq. we get,

(48 - x - \frac { (702 - x^2)}{x} ) \times (48 - x ) = 124

on solving for x , we putting the value in y = \frac { (702- x^2)}{x} we get, y^2 = \frac { 459073}{576} => y = 791.57 .