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

enter image description here

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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – robjohn
    Apr 3, 2022 at 1:29

2 Answers 2


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.

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


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:


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:

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


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