For reference (exact copy of the question):

In the acute triangle $ABC$, the distance between the feet of the relative heights to sides $AB$ and $BC$ is $24$. Calculate the measure of the circumradius of triangle $ABC$. $\angle B = 37^\circ$


My progress:

My figure and the relationships I found

enter image description here

I tried to draw $DH\perp AC$ in $H$ $\implies \triangle DCH$ is notable ($37^\circ:53^\circ$) therefore sides = $3k:4k:5k$

$FE$ is a right triangle cevian...but I can't see where it will go into the solution.

  • $\begingroup$ @ACB..Translator error..the corrector would be cevian..In geometry, a cevian is a line that intersects both a triangle's vertex, and also the side that is opposite to that vertex. $\endgroup$ Oct 10, 2021 at 21:59
  • $\begingroup$ I know cevian. But why right triangle cevian? $\endgroup$
    – ACB
    Oct 11, 2021 at 4:13
  • $\begingroup$ @ACB .. I meant $FE$ is cevian from the right triangle $BFC$ $\endgroup$ Oct 11, 2021 at 10:21

4 Answers 4


enter image description here

Here is an approach that avoids trigonometry. In $\triangle ADH$, $\frac{AH}{AD} = \frac{3}{5} \implies AD = R = \frac{5}{3} AH$.

As $AH = AC/2$, $R = \frac{5}{6} AC \tag1$

$H$ is the circumcenter of right triangles $\triangle AFC$ and $\triangle AEC$.

So, $FH = EH = AC/2$

$\angle AHF = 180^\circ - 2 \angle A, \angle EHC = 180^\circ - 2 \angle C$

That leads to $\angle EHF = 180^\circ - 2 \angle B$

As $\triangle EHF$ is isosceles and $M$ is the foot of perpendicular from $H$ to $FE$,

$\angle HFM = \angle HEM = \angle B$ and $FM = ME = 12$.

In $\triangle FHM$, $ \displaystyle \frac{FM}{FH} = \frac{4}{5}$

$FH = 15 = AC/2 \implies AC = 30$

Using $(1)$, $R = 25$


enter image description here

Let $BG$ be the altitude from $B$ to $AC$. $\triangle EFG$ is orthic triangle.

A property of orthic triangle: Each line from the circumcenter of the parent triangle to a vertex is always perpendicular to the corresponding side of the orthic triangle.

As $D$ is the circumcenter of $\triangle ABC$, $ ~ CD \perp EG ~ $ and $AD \perp FG$.

So, $\angle EGF = 180^\circ - \angle ADC = 106^\circ$

If $R_O$ is the circumradius of $\triangle EGF$, using extended sine law in $\triangle EGF, EF = 24 = 2 R_O \sin \angle EGF$

$\implies \displaystyle R_O = \frac{12}{\sin 106^\circ} \tag1$

$ \sin 106^\circ = 2 \sin 53^\circ \cos 53^\circ = \displaystyle 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$

So from $(1)$, $ \displaystyle R_O = \frac{25}{2}$

Now we use another property of orthic triangle: Circumradius of orthic triangle is half the circumradius of the parent triangle.

$\therefore R = 2 R_O = 25$

  • $\begingroup$ A link that has more details on both the properties of orthic triangle I used: mathworld.wolfram.com/OrthicTriangle.html $\endgroup$
    – Math Lover
    Oct 10, 2021 at 17:30
  • $\begingroup$ thanfk for a link and solution... $\endgroup$ Oct 10, 2021 at 22:16
  • 1
    $\begingroup$ I believe there is an alternative without using trigonometry...but it is not easy to visualize $\endgroup$ Oct 10, 2021 at 22:18
  • $\begingroup$ @petaarantes you are right. I couldn't think of one yesterday night.. $\endgroup$
    – Math Lover
    Oct 11, 2021 at 10:35
  • 1
    $\begingroup$ @ACB I wonder that as well, now :) I was more decided to use the two mentioned properties of the orthic triangle I guess $\endgroup$
    – Math Lover
    Oct 11, 2021 at 10:54

orthic triangle

Property: In a triangle $ABC$, the side lengths of the orthic triangle are given by

$\begin{align}a'&=a\cos A\\b'&=b\cos B\\c'&=c\cos C\end{align}$

Accordingly, in your image,


$EF=AC\cos B$.

As we know $EF=24$ (given) and $\cos B=\cos 37^\circ=\frac45$, this implies $AC=30$.

Therefore $CH=\displaystyle{\frac{AC}2}=15$.

From the special right triangle $DCH$, we find $DC=\frac53CH=25$. $$\therefore R=25$$

  • 1
    $\begingroup$ great...thanks for solution $\endgroup$ Oct 11, 2021 at 10:22

Let altitudes $AE$ and $CF$ intersect at orthocenter $J$.

Since $\angle BEJ + \angle JFB = 90^\circ + 90^\circ = 180^\circ$, $BFJE$ is a cyclic quadrilateral with diameter $BJ$, by sine rule in $\triangle BFE$,

$$ \frac{EF}{\sin B} = BJ \tag{1}$$

Denote circumradius of $\triangle ABC$ by $R$. Then in terms of $R$ we have formula for $BJ$ as $$BJ = 2DH = 2R\cos B \tag{2}$$

Combining both equations, one gets $\boxed{R=25}$.

  • $\begingroup$ i didn't undertand....by sine rule in △BFE,...? $ \frac{EF}{sin B} = BJ, but △ BFE$ isn't right triangle... $\endgroup$ Oct 10, 2021 at 23:06
  • $\begingroup$ @petaarantes Recall sine rule says $a/ \sin A= 2r$ where $2r$ is circumdiameter. And we have shown that $BJ$ is diameter of circle passing through $B, E, F$. $\endgroup$ Oct 11, 2021 at 3:48
  • $\begingroup$ Ah...yes...I thought you were using the opposite side rule with the hypotenuse...Now I get it...thanks for the clarification $\endgroup$ Oct 11, 2021 at 10:15

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