Htykuut

A circle is inscribed in trapezoid $ABCD$. Let $K,L,M,N$ be the points of intersection of the circle with diagonals $AC$ and $BD$ respectively. $K$ is between $A$ and $L$ and $M$ is between $B$ and $N$. Given that $AK*LC = 16$ and $BM*ND = frac{9}{4}$, find the radius of the circle

I was able to deduce a few elementary things

$AB + CD = AD + BC$ and also tried using power of point and use the given products but didnt get anything useful

When all else fails, get a bigger hammer

...which I did and found that the answer was $R=6$.

I won't explain why $AC$, $BD$ $PR$ and $QS$ concur at $F$. It's just a special case of
Brianchon's theorem. So let's start from there. Notice the angles $angle CAB=angle ACD=alpha$, $angle ABD=angle BDC=beta$ and segment $OF=x$. We'll use them all the time. For the sake of simplicity I will also introduce the following symbols: $AK=p_1$, $LC=p_2$, $BM=q_1$, $DN=q_2$. We know that $p_1p_2=16$ and $q_1q_2=9/4$.

First, let's try to find $p_1=AK$. The same approach will be used to find segments $p_2,q_1,q_2$

Take a look at quadrilateral $AQOK$:

$$AQ=FQcotalpha=AKcos alpha+OKsinvarphi$$

$$QO=AKsinalpha+OKcosvarphi$$

This leads to the following equations:

$$p_1cosalpha+Rsinvarphi=(R+x)cotalpha$$
$$p_1sinalpha+Rcosvarphi=R$$

or:

$$Rsinvarphi=(R+x)cotalpha-p_1cosalpha$$
$$Rcosvarphi=R-p_1sinalpha$$

Square these two equations and add them. This will eliminate the angle $varphi$ that we don't care about. You will end up with a quadratic equation with respect to $p_1$:

$$R^2=((R+x)cotalpha-p_1cosalpha)^2+(R-p_1sinalpha)^2$$

This equation has two solutions. The bigger one is actually equal to $AL$ and the smaller one is $AK=p_1$:

$$p_1=-sqrt{R^2-frac{1}{2} x^2 (1+cos (2 alpha))}+(R+x)cos alpha cot alpha +R sin alphatag{1}$$

There is no need to calculate $p_2$ in a similar way. Everything would look the same, except that we would use $R-x$ instead of $R+x$ along the way. So we can obtain an expression for $p_2$ just by replacing $x$ with $-x$:

$$p_2=-sqrt{R^2-frac{1}{2} x^2 (1+cos (2 alpha))}+(R-x)cos alpha cot alpha +R sin alphatag{2}$$

Now comes the hardest part: you have to multiply (1) and (2). It looks like we are going nowhere... After a lot of simplifications, we get:

$$p_1p_2=-frac{1}{2} csc ^2alpha left(left(R^2+x^2right)cos (2 alpha) +2 R sin alpha sqrt{4 R^2-2 x^2 (1+cos (2 alpha))}-3 R^2+x^2right)tag{3}$$

At least we know it's 16 :)

The good thing is that we don't have to go through the same torture to find $q_1q_2$, the process is exactly the same except that we have $beta$ instead of $alpha$:

$$q_1q_2=-frac{1}{2} csc ^2beta left(left(R^2+x^2right)cos (2 beta) +2 R sin beta sqrt{4 R^2-2 x^2 (1+cos (2 beta))}-3 R^2+x^2right)tag{4}$$

Like it's not difficult enough we have to introduce the fact that $x$, $alpha$ and $beta$ are not independent! There's a connection between them.

Take a look at triangle $triangle ABC$:

$$AB=AQ+QB=(R+x)(cotalpha+cotbeta)$$

$$BC=BR+RC=BQ+CS=(R+x)cotbeta+(R-x)cotalpha$$

$$AC={2R over sinalpha}$$

If you replace all that into a well-known equation:

$$BC^2=AB^2+AC^2-2 ABcdot ACcosalpha$$

...you will get a very simple relation (trust me):

$$x^2=R^2(1-tanalphatanbeta)tag{5}$$

Now replace (5) into (3) and (4) and we'll get something that we can easily deal with:

$$p_1p_2=R^2left(2+frac{tanbeta}{tanalpha}-2sqrt{1+frac{tanbeta}{tanalpha}}right)$$

$$q_1q_2=R^2left(2+frac{tanalpha}{tanbeta}-2sqrt{1+frac{tanalpha}{tanbeta}}right)$$

...or, if we introduce $z=frac{tanalpha}{tanbeta}$:

$$p_1p_2=R^2(2+z-2sqrt{1+z})=R^2(sqrt{1+z}-1)^2=16tag{6}$$

$$q_1q_2=R^2left(2+frac1z-2sqrt{1+frac1z}right)=R^2left(sqrt{1+frac1z}-1right)^2=frac{9}{4}tag{7}$$

Now divide (6) by (7) and you get:

$$left(frac{sqrt{1+z}-1}{sqrt{1+frac1z}-1}right)^2=frac{p_1p_2}{q_1q_2}=frac{16}{frac94}=frac{64}{9}$$

$$frac{sqrt{1+z}-1}{sqrt{1+frac1z}-1}=frac{8}{3}$$

$$3(sqrt{1+z}-1)=8left(sqrt{1+frac1z}-1right)tag{8}$$

You should be able to show that equation (8) has only one solution:

$$z=frac{16}9$$

Replace that into (6) and you get:

$$R^2=frac{p_1p_2}{(sqrt{1+z}-1)^2}=36implies R=6$$