1982 AHSME Problems/Problem 24

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Solution 1

We now know that because $AG=2, GF=13,$ and $FC=1$ the whole perimeter of the triangle. And because $2+13+1=16,$ the equilateral triangle will have side length $16$. Harnessing the power of point $C$, we have $(CF)(CG)=(CE)(CD).$ We also have on point $A$: $(AH)(AJ)=(AG)(AF).$ We will simplify this, by inputting values we already know.

What we know:

Side length $16$

$AG=2, GF=13,$ and $FC=1$

Now we know what $CF$ is, $1.$ We also can derive $CG,$ as it equals $CF+FG=13+1=14.$

$(CF)(CG)=(CE)(CD),$ so by the prior calculations, $(1)(13)=(CE)(CB).$ Therefore $13=(CE)(CD).$

We can update our chart:

What we know:

Side length 16

$AG=2, GF=13,$ and $FC=1$

$13=(CE)(CD).$

$CG=14$


$(AH)(AJ)=(AG)(AF),$ and it's pretty obvious that $AG=2.$ We can also derive $AF, 2+GF=15.$ Therefore $(AH)(AJ)=15.$ However, we know the value of $HJ!$ Because $HJ=AJ-AH,$ we have $AJ=7+AH.$ Therefore $(AH)(7+AH)=15.$ We can distribute: $AH^2+7AH-15=0,$ and $AH = \sqrt{109}/2 - 7/2.$ But, we know that the side lengths must sum to $16,$ so we have $BJ=16-(AH+HJ)=25/2 - \sqrt{109}/2.$ We can now update our 'What we Know section.

What we know:

Side length 16

$AG=2, GF=13,$ and $FC=1$

$13=(CE)(CD).$

$CG=14$

$AH = \sqrt{109}/2 - 7/2.$

$BJ=25/2 - \sqrt{109}/2.$

Taking the power of point B, we have $(BD)(BE)=(BJ)(BH,)$ and because we already know $AH$, we know $BH=16-AH=39/2 - \sqrt{109}/2.$ Now we shall multiply $BH$ and $BJ$ to get $(25/2 - \sqrt{109}/2) \cdot (39/2 - \sqrt{109}/2)= 271 - 16 \sqrt{109}.$ Therefore $BD(BE)=271 - 16 \sqrt{109}.$

Updating our chart,

What we know:

Side length 16

$AG=2, GF=13,$ and $FC=1$

$13=(CE)(CD).$

$CG=14$

$AH = \sqrt{109}/2 - 7/2.$

$BJ=25/2 - \sqrt{109}/2.$

$BD(BE)=271 - 16 \sqrt{109}.$

$HJ=7$

$BH=39/2 - \sqrt{109}/2.$

$CE=16-BE$

$CD=16-BD$

We can now substitute and have 2 equations and two variables.$CE=16-BE$ and $CD=16-BD$ are especially useful. Because $13=(CE)(CD)=(16-BE)(16-BD),$ and $BD(BE)=271 - 16 \sqrt{109},$ we have $BD= 15, 3, 29,$ or $17,$ and $BE=15, 3, 29, or 17.$ However, because $BD<BE,$ we have $BD=3$ or $17$ and $BE= 15$ or $29.$ Because the differences $29-17$ and $15-3$ are constantly $12,$ and $BD-BE=DE, DE=12!$