What's the measure of the circumradius of triangle ABC?

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 $$ABC$$. $$\angle B = 37^\circ$$

(Answer:$$25$$}

My progress:

My figure and the relationships I found

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.

• @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. Oct 10, 2021 at 21:59
• I know cevian. But why right triangle cevian?
– ACB
Oct 11, 2021 at 4:13
• @ACB .. I meant $FE$ is cevian from the right triangle $BFC$ Oct 11, 2021 at 10:21

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

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.

$$\therefore R = 2 R_O = 25$$

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

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}

$$EF=AC\cos B$$.

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 $$DC=\frac53CH=25$$. $$\therefore R=25$$

• great...thanks for solution 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 $$\boxed{R=25}$$.

• i didn't undertand....by sine rule in △BFE,...? $\frac{EF}{sin B} = BJ, but △ BFE$ isn't right triangle... Oct 10, 2021 at 23:06
• @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 $B, E, F$. Oct 11, 2021 at 3:48
• Ah...yes...I thought you were using the opposite side rule with the hypotenuse...Now I get it...thanks for the clarification Oct 11, 2021 at 10:15