# Triangle with side lengths and three tangent circles, finding ratio of lengths $KF/KE$

Triangle $$ABC$$ has side lengths $$AB=7, BC=8,$$ and $$CA=9.$$ Its incircle $$\Gamma$$ meets sides $$BC,CA,$$ and $$AB$$ at $$D,E,F$$ respectively. Let $$AD$$ intersect $$\Gamma$$ at a point $$P \neq D.$$ The circle passing through $$A$$ and $$P$$ tangent to $$\Gamma$$ intersects the circle passing through $$A$$ and $$D$$ tangent to $$\Gamma$$ at a point $$K \neq A.$$ Find $$\tfrac{KF}{KE}.$$

I was not sure how to approach this problem and am still not completely sure. First, I drew an almost to scale diagram (it wasn't 100% to scale because I was not sure how to perfectly construct the circle passing through $$A$$ and $$D$$ tangent to $$\Gamma.$$ (It would be helpful for a strategy to precisely construct the circle passing through $$A$$ and $$D$$ tangent to $$\Gamma$$ though). The diagram is shown below By some approximation, the ratio was around $$37/42$$ but I'm not sure if this is correct, especially because it is an unrigorous approximation. I have an idea to set coordinates and bash out the coordinates for $$K,F,E$$ but this would be tedious and an unclean strategy. May I have some help? Thanks in advance.

• I used Mathematica to draw this diagram and as a by product, I got that $\overline{KF}=36\sqrt{\frac{15}{1687}}$ and $\overline{KE}=20\sqrt{\frac{35}{723}}$. The ratio of these lengths is $\frac{27}{35}$. I will see if I can come up with a justification for this.
– robjohn
Apr 3, 2022 at 1:29

Let the circle through $$A$$ and $$D$$ tangent to $$BC$$ be $$\omega_1$$ and the circle through $$A$$ and $$P$$ tangent to $$\Gamma$$ be $$\omega_2$$. Then by radical axes on $$\omega_1$$, $$\omega_2$$ and $$\Gamma$$, we see that $$AK$$, $$BC$$, and the common tangent of $$\omega_2$$ and $$\Gamma$$ are concurrent at a point, say $$X$$. Now since $$PEDF$$ is a harmonic quadrilateral, we see that $$EF$$ passes through $$X$$, so $$XE \cdot XF = XD^2 = XK \cdot XA$$, and consequently $$AKFE$$ is cyclic. Now let $$\omega_1$$ intersect $$AC$$ and $$AB$$ at $$Q$$ and $$R$$ respectively. Then by spiral similarity, $$\triangle KEQ \sim \triangle KFR$$, so $$\frac{KF}{KE} = \frac{FR}{EQ}$$, and all there is left is to compute these two lengths. The side lengths of the triangle are $$a = 8$$, $$b = 9$$ and $$c = 7$$, and the semiperimeter $$s = 12$$. Since $$D$$, $$E$$, $$F$$ are the touchpoints of the incircle with the sides, we have $$BD = BF = s - b = 12 - 9 = 3$$, and $$CD = CE = s - c = 12 - 7 = 5$$. Therefore, $$EQ = CE - CQ = CD - \frac{CD^2}{CA} = 5 - \frac{25}{9} = \frac{20}{9}$$ and similarly, $$FR = BF - BR = BD - \frac{BD^2}{BA} = 3 - \frac{9}{7} = \frac{12}{7}.$$ Hence finally, $$\frac{KF}{KE} = \frac{FR}{EQ} = \frac{12}{7} \cdot \frac{9}{20} = \frac{27}{35}.$$

Note: You should post problems like these to https://artofproblemsolving.com. There, you are more likely to get an olympiad-flavoured answer.

• Thank you for the response, your solution is very fascinating! I will be sure to check out artofproblemsolving Apr 2, 2022 at 15:08

Hints:

1- find measure of $$\angle BAC\approx 58.4^o$$ from the measures of the sides of triangle ABC.

2- find the measure of radius of circle $$\Gamma$$ from this relation:

$$r=\sqrt{\frac{(p-a)(p-b)(p-c)}p}$$

where $$p=\frac{a+b+c}2$$, you must get $$r\approx 2,24$$

3- find $$AE$$ it must be $$4$$

4-Extend KF to touch circle $$Gamma$$ at point H. show that:

• $$\angle HKE=\angle FKE=\approx 58,4^o$$

• $$KH=KE$$ and $$KF=KJ$$, where J is intersection of circle $$Gamma$$ with KE.we have $$KH=KE\approx KG$$

5- Draw a tangent from K to circle $$\Gamma$$ to touch it at point I.

6-Connect K to G , $$\angle GKI=\frac {58.4}2=29.2^o$$ and you have r , find measure of KI=4.Also calculate the measure of KG, it is about $$4.4$$.

Now use this relation:

$$(KF=KJ)\times (KE=KG=4.4)=KI^2$$

you get:

$$\frac {KF}{KE}=\frac{3.4}{4.4}=\approx 0.77$$