https://doi.org/10.53656/math2026-4-1-ptr

Abstract. A property of a tetrahedron with a right trihedral angle is considered. Several approaches to its proof are presented. For this purpose, auxiliary statements are discussed, which also have independent significance.

Keywords: stereometry; right trihedral angle; tetrahedron; Lagrange’s formula; Euler’s formula; de Gua’s theorem; mathematical olympiad

1. Introduction

The idea for this article originates from Problem 11.3 of the Bulgarian Spring Mathematical Competition 2026, proposed by the second author. The problem is stereometric, which is something refreshing. Problems of this type have been deliberately neglected in recent decades because they are excluded from the content framework of the International Mathematical Olympiad (IMO). We hope that this will change after 2027, which would lead to the inclusion of new and interesting ideas.

Let us consider an acute-angled \(\triangle A B C\). Let \(O\) and \(H\) be respectively the circumcenter of \(\triangle A B C\) and the orthocenter of \(\triangle A B C\). We want to find a point \(X\) in space such that the trihedral angle at the vertex \(X\) of \(X A B C\) is right. There are two such points, symmetric with respect to the plane determined by \(A B C\). It turns out that for such a point \(X\), the following relation holds

(1) \[ X H^{2}+X O^{2}=R^{2} \]

where \(R\) is the radius of the circumcircle of \(\triangle A B C\). This interesting fact is in fact the mentioned problem 11.3 from the Spring math competition 2026. Below we present four different approaches to proving (1). Before that, auxiliary statements are considered, which also have independent significance – Lagrange’s formula, Euler’s formula, and de Gua’s theorem, etc.

The article is intended for students actively engaged in Olympiad mathematics, as well as for teachers and coaches who train them.

2. Some useful facts

Many of these results are standard in olympiad geometry; see, e.g. (Prasolov, 2001).

Proposition 1. (Lagrange’s formula) Let \(A_{1}, A_{2}, \ldots, A_{n}\) be points in space, where the point \(A_{i}\) has mass \(m_{i}, 1 \leq i \leq n\). Let \(G\) be the center of mass of these points. Then for an arbitrary point \(O\) in space we have

(2) \[ \sum_{i=1}^{n} m_{i} O A_{i}^{2}=\left(\sum_{i=1}^{n} m_{i}\right) O G^{2}+\sum_{i=1}^{n} m_{i} G A_{i}^{2} \]

Before proceeding to the proof, let us note that the above formula has applications in mechanics, and in geometry it is mainly used in the special case \(m_{i}=1,1 \leq i \leq n\).

Proof. Since \(G\) is the center of mass of the points \(A_{1}, \ldots A_{n}\), we have

(3) \[ \sum_{i=1}^{n} m_{i} \overrightarrow{G A_{i}}=0 \]

From \(\overrightarrow{O A_{i}}=\overrightarrow{O G}+\overrightarrow{G A_{i}}\), after squaring (dot product) we obtain

\[ m_{i} O A_{i}^{2}=m_{i} O G^{2}+2 m_{i} \overrightarrow{O G} \cdot \overrightarrow{G A_{i}}+m_{i} G A_{i}^{2} \] Summing these equalities for \(i=1,2, \ldots, n\) and using (3), the result follows.

Let us consider two simple consequences of the above statement.

Corollary 1. (median formula) For \(\triangle A B C\) with midpoint \(M\) of \(B C\) we have

\[ 2 A M^{2}=A B^{2}+A C^{2}-\tfrac{B C^{2}}{2} \] Proof. We apply Lagrange’s formula for \(O \equiv A\), points \(B, C\) and \(m_{1}=m_{2}=\) 1. Since \(M\) is the center of mass of the points \(B, C\), we obtain

\[ A B^{2}+A C^{2}=2 A M^{2}+M B^{2}+M C^{2}=2 A M^{2}+\tfrac{B C^{2}}{2} \] Corollary 2. Let \(A_{1}, A_{2}, \ldots, A_{n}\) be points in space, and let \(G\) be their center of mass. Then the set of points \(X\) for which

\[ \sum_{i=1}^{n} X A_{i}^{2}=C \]

where \(C \gt 0\) is a constant, is either empty or a sphere with center at the point \(G\).

Proof. Indeed, applying Lagrange’s formula for \(m_{i}=1\), we obtain

\[ C=\sum_{i=1}^{n} X A_{i}^{2}=n X G^{2}+\sum_{i=1}^{n} G A_{i}^{2} \]

from which, \(X G^{2}=C-\sum_{i=1}^{n} G A_{i}^{2}\). Note that the expression on the right-hand side depends only on the points \(A_{i}\), which shows that for \(C \geq \sum_{i=1}^{n} G A_{i}^{2}\), the distance \(X G\) is constant. The converse also follows directly.

Example 1. Let \(A B C\) be a triangle with orthocenter \(H\) and circumcenter \(O\), and let \(R\) be its circumradius. We seek to determine all points \(X\) in space such that

\[ X O^{2}+X H^{2}=R^{2} \]

Let \(K\) denote the center of the nine-point (Feuerbach) circle, i.e., the midpoint of \(O H\). For further properties of this circle, see (Coxeter & Greitzer, 1967).

As we established, the locus of points \(X\) satisfying a relation of the form \(X O^{2}+X H^{2}=\) const is either empty or a sphere. Applying Lagrange’s formula, we obtain

\[ R^{2}=X O^{2}+X H^{2}=2 X K^{2}+K O^{2}+K H^{2}=2 X K^{2}+\tfrac{1}{2} O H^{2} \] and hence

\[ X K=\tfrac{1}{2} \sqrt{2 R^{2}-O H^{2}} \]

Thus:

• if \(2 R^{2}-O H^{2} \gt 0\), the locus of \(X\) is a sphere with center \(K\) and radius \(\tfrac{1}{2} \sqrt{2 R^{2}-O H^{2}}\);

• if \(2 R^{2}-O H^{2}=0\), the locus consists of the single point \(K\);

• if \(2 R^{2}-O H^{2} \lt 0\), no such point \(X\) exists.

A straightforward computation yields

(4) \[ 2 R^{2}-O H^{2}=R^{2}(1+8 \cos A \cos B \cos C) \]

Hence, the nature of the locus is determined by the sign of the right-hand side of (4) . In particular, if \(\triangle A B C\) is acute, then \(2 R^{2}-O H^{2} \gt 0\), whereas for sufficiently obtuse triangles one has \(2 R^{2}-O H^{2} \lt 0\).

In the acute case, (4) further implies

\[ X K=\tfrac{1}{2} \sqrt{2 R^{2}-O H^{2}} \gt \tfrac{R}{2} \] Therefore, the locus of \(X\) is a sphere centered at \(K\) that strictly contains the nine-point circle, whose radius is R/2, see (Coxeter & Greitzer, 1967).

Proposition 2. (Euler’s formula, see (Johnson, 2007)) For \(\triangle A B C\) with orthocenter \(H\) and circumradius \(R\) we have

(5) \[ O H^{2}=9 R^{2}-a^{2}-b^{2}-c^{2} \]

Proof. Let \(G\) be the centroid of \(\triangle A B C\). From Lagrange’s formula (2), applied to the points \(O, A, B, C\), we obtain

\[ O A^{2}+O B^{2}+O C^{2}=3 O G^{2}+G A^{2}+G B^{2}+G C^{2} \] which gives

(6) \[ 3 R^{2}=3 O G^{2}+G A^{2}+G B^{2}+G C^{2} . \]

From \(\overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C}=0\), after squaring we have

(7) \[ G A^{2}+G B^{2}+G C^{2}+2(\overrightarrow{G A} \cdot \overrightarrow{G B}+\overrightarrow{G A} \cdot \overrightarrow{G C}+\overrightarrow{G B} \cdot \overrightarrow{G C})=0 . \]

On the other hand,

\[ \begin{aligned} & a^{2}=(\overrightarrow{G C}-\overrightarrow{G B})^{2}=G B^{2}+G C^{2}-2 \overrightarrow{G B} \cdot \overrightarrow{G C}, \\ & b^{2}=(\overrightarrow{G A}-\overrightarrow{G C})^{2}=G A^{2}+G C^{2}-2 \overrightarrow{G A} \cdot \overrightarrow{G C}, \\ & c^{2}=(\overrightarrow{G B}-\overrightarrow{G A})^{2}=G B^{2}+G A^{2}-2 \overrightarrow{G A} \cdot \overrightarrow{G B} . \end{aligned} \] Summing the above equalities and using (7), we obtain \[ 3\left(G A^{2}+G B^{2}+G C^{2}\right)=a^{2}+b^{2}+c^{2} . \] Substituting into (6) we get

\[ 9 O G^{2}=9 R^{2}-a^{2}-b^{2}-c^{2} \] Finally, since the points \(O, G, H\) lie on a line in this order (Euler line) and \(O H=3 O G\), we obtain

\[ O H^{2}=9 R^{2}-a^{2}-b^{2}-c^{2} . \]

Now we present a well-known formula for the length of the segment connecting a vertex of a triangle with its orthocenter.

Proposition 3. Let \(A B C\) be an acute triangle with orthocenter \(H\) and circumradius \(R\). Then

\[ A H^{2}=4 R^{2}-B C^{2} . \] Proof. Let \(H^{\prime}\) be the reflection of \(H\) with respect to \(A C\). It is known that \(H^{\prime}\) lies on the circumcircle. We have \(A H=A H^{\prime}\). Let \(H^{\prime} O\) intersect the circumcircle again at the point \(H^{\prime \prime}\). Since \(H^{\prime} H^{\prime \prime}\) is a diameter, it is easy to see that \(\angle H^{\prime \prime} H^{\prime} A=\angle A\), i.e. \(A H^{\prime \prime}=B C\). From the Pythagorean theorem in \(\triangle A H^{\prime} H^{\prime \prime}\) we obtain

\[ 4 R^{2}=H^{\prime} H^{\prime \prime 2}=A H^{\prime 2}+A H^{\prime \prime 2}=A H^{2}+B C^{2} \]

The following result can be viewed as a three-dimensional analogue of the Pythagorean theorem. To the best of the authors’ knowledge, this orthocenterbased proof of de Gua’s theorem does not appear in the literature

Proposition 4. (de Gua’s theorem) Let \(A B C D\) be a tetrahedron with a right trihedral angle at the vertex \(D\). Then for the faces we have

\[ S_{A B C}^{2}=S_{D A B}^{2}+S_{D B C}^{2}+S_{D A C}^{2} \] Proof. Let \(H\) be the orthocenter of \(\triangle A B C\). From the theorem of the three perpendiculars it follows that \(D\) projects onto the plane of \(A B C\) at the point \(H\). Denote by \(A_{1}, B_{1}, C_{1}\) the feet of the perpendiculars from the vertices \(A, B, C\) to the sides of the triangle, respectively. Note that the triangles \(\triangle A D A_{1}, B D B_{1}, C D C_{1}\) are right-angled at \(D\), and \(D H\) is an altitude for each of them. From this it follows

\[ D A_{1}^{2}=A_{1} H \cdot A_{1} A ; D B_{1}^{2}=B_{1} H \cdot B_{1} B ; D C_{1}^{2}=C_{1} H \cdot C_{1} C \]

Since \(D A_{1} \perp B C, D B_{1} \perp A C, D C_{1} \perp A B\), DB1 ⊥ AC, DC1 ⊥ AB, we obtain

\[ 4 S_{D B C}^{2}=D A_{1}^{2} \cdot B C^{2}=A_{1} H \cdot B C \cdot A A_{1} \cdot B C=4 S_{H B C} \cdot S_{A B C} \] Similarly, \(4 S_{D A C}^{2}=4 S_{H A C} \cdot S_{A B C}\) and \(4 S_{D A B}^{2}=4 S_{H A B} \cdot S_{A B C}\). Summing these equalities, we obtain

\[ S_{D A B}^{2}+S_{D B C}^{2}+S_{D A C}^{2}=S_{A B C}\left(S_{H A B}+S_{H B C}+S_{H A C}\right)=S_{A B C}^{2} \]

3. Applications

Let us now consider the following problem.

Problem 1. (Spring competition 2026, Problem 11.3) Let ABCD be a tetrahedron with a right trihedral angle at the vertex \(D\). The point \(H\) is the orthocenter of \(\triangle A B C\), and the circumcircle of \(\triangle A B C\) has center \(O\) and radius \(R\). Prove that \(D O^{2}+D H^{2}=R^{2}\).

We will prove the statement in 4 different ways. Let us begin with the most elegant one, in our opinion. It was found by two students during the competition.

First solution. Let \(H_{1}\) be the foot of the perpendicular from \(A\) to \(B C\), and let \(H_{1}^{\prime}\) be the second intersection point of \(A H_{1}\) with the circle \(k\), see Fig. 1 and Fig. 2. We have

(8) \[ \begin{aligned} & D O^{2}+D H^{2}=R^{2} \\ & \quad \Longleftrightarrow D H^{2}+O H^{2}+D H^{2}=R^{2} \\ & \Longleftrightarrow 2 D H^{2}=R^{2}-O H^{2}=(R-O H)(R+O H) \end{aligned} \]

Note that the right-hand side equals the power of point \(H\) with respect to the circle \(k\) (Coxeter & Greitzer, 1967). This means \((R-O H)(R+O H)=\) \(A H \cdot H H_{1}^{\prime}\). Moreover, it is known that \(H H_{1}=H_{1} H_{1}^{\prime}\). Therefore

\[ (R-O H)(R+O H)=A H \cdot H H_{1}^{\prime}=2 A H \cdot H H_{1} . \] From (8) we obtain

\[ D O^{2}+D H^{2}=R^{2} \Longleftrightarrow D H^{2}=A H \cdot H H_{1} . \] Since \(\triangle D A H_{1}\) is right-angled and \(D H\) is the altitude to \(A H_{1}\), it follows that \(D H^{2}=A H \cdot H H_{1}\). Hence \(D O^{2}+D H^{2}=R^{2}\).

Second solution. Let \(a=B C, b=C A, c=A B\). It is easy to see that

\(D H \perp(A B C)\) (e.g. by the theorem of the three perpendiculars). Then

(9) \[ D H^{2}=D A^{2}-A H^{2}=D A^{2}-4 R^{2}+a^{2}=D B^{2}-4 R^{2}+b^{2}=D C^{2}-4 R^{2}+c^{2} \] and since \(D B^{2}+D C^{2}=a^{2}, D C^{2}+D A^{2}=b^{2}, D A^{2}+D B^{2}=c^{2}\), we get \(6 D H^{2}=2\left(D A^{2}+D B^{2}+D C^{2}+a^{2}+b^{2}+c^{2}-12 R^{2}\right)=3\left(a^{2}+b^{2}+c^{2}-8 R^{2}\right)\). From this and the Euler’s formula, \(O H^{2}=9 R^{2}-a^{2}-b^{2}-c^{2}\), it follows that \[ D O^{2}+D H^{2}=O H^{2}+2 D H^{2}=R^{2} . \]

Third solution. We introduce a coordinate system with origin at \(D\) and axes along the edges \(D A, D B, D C\). Let \(A=\left(a_{1}, 0,0\right), B=\left(0, a_{2}, 0\right), C=\left(0,0, a_{3}\right)\) and \(O=\left(o_{1}, o_{2}, o_{3}\right)\). The plane (\(A B C\) ) has equation \(x_{1} / a_{1}+x_{2} / a_{2}+x_{3} / a_{3}=1\) and \(H\) is the projection of \(D\) onto it. Using the formula for the distance from

a point to a plane, we obtain

(10) \[ D H=\left(a_{1}^{-2}+a_{2}^{-2}+a_{3}^{-2}\right)^{-1 / 2} . \]

Further, since \(O \in(A B C)\), we have

(11) \[ \tfrac{o_{1}}{a_{1}}+\tfrac{o_{2}}{a_{2}}+\tfrac{o_{3}}{a_{3}}=1 . \]

Moreover,

\[ \left(o_{1}-a_{1}\right)^{2}+o_{2}^{2}+o_{3}^{2}=o_{1}^{2}+\left(o_{2}-a_{2}\right)^{2}+o_{3}^{2}=o_{1}^{2}+o_{2}^{2}+\left(o_{3}-a_{3}\right)^{2}=R^{2} . \] Since, \(D O^{2}=o_{1}^{2}+o_{2}^{2}+o_{3}^{2}\), we obtain \[ a_{i}^{2}-2 o_{i} a_{i}=R^{2}-D O^{2}, \text { i.e. } 1-2 \tfrac{o_{i}}{a_{i}}=\tfrac{R^{2}-D O^{2}}{a_{i}^{2}}, \quad i=1,2,3 . \] It remains to sum these three equalities and use (10) and (11).

Fourth solution. Let \(x=D A, y=D B, z=D C\) and \(S=S_{A B C}\). The idea is to express \(D H^{2}, R^{2}\) and \(D O^{2}\) in terms of \(x, y, z\). From the theorem of the three perpendiculars it follows that \(D H \perp(A B C)\). Writing the volume of \(A B C D\) in two ways, we obtain

(12) \[ D H=\tfrac{x y z}{2 S} . \]

From \(R=\tfrac{a . b . c}{4 S}\) we have

(13) \[ R^{2}=\tfrac{\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)\left(x^{2}+z^{2}\right)}{16 S^{2}} . \]

Let \(G\) denote the centroid of \(\triangle A B C\). We have

\[ \overrightarrow{D G}=\tfrac{1}{3}(\overrightarrow{D A}+\overrightarrow{D B}+\overrightarrow{D C}), \]

from which \[ \begin{gathered} D G^{2}=\tfrac{1}{9}\left(x^{2}+y^{2}+z^{2}\right) \\ G H^{2}=D G^{2}-D H^{2}=\tfrac{1}{9}\left(x^{2}+y^{2}+z^{2}\right)-\tfrac{x^{2} y^{2} z^{2}}{4 S^{2}} \end{gathered} \] Since \(O, G\) and \(H\) lie on the Euler line and \(O G: G H=1: 2\), we have \(H O=\tfrac{3}{2} G H\) and

\[ H O^{2}=\tfrac{9}{4} H G^{2}=\tfrac{1}{4}\left(x^{2}+y^{2}+z^{2}\right)-\tfrac{9}{16} \tfrac{x^{2} y^{2} z^{2}}{S^{2}} . \] From \(D O^{2}=D H^{2}+H O^{2}\) and (12) we have

\[ \begin{gathered} D O^{2}=\tfrac{x^{2} y^{2} z^{2}}{4 S^{2}}+\tfrac{1}{4}\left(x^{2}+y^{2}+z^{2}\right)-\tfrac{9}{16} \tfrac{x^{2} y^{2} z^{2}}{S^{2}} \\ D O^{2}=\tfrac{1}{4}\left(x^{2}+y^{2}+z^{2}\right)-\tfrac{5}{16} \tfrac{x^{2} y^{2} z^{2}}{S^{2}} \\ D O^{2}+D H^{2}=\tfrac{1}{4}\left(x^{2}+y^{2}+z^{2}\right)-\tfrac{5}{16} \tfrac{x^{2} y^{2} z^{2}}{S^{2}}+\tfrac{x^{2} y^{2} z^{2}}{4 S^{2}}=\tfrac{1}{4}\left(x^{2}+y^{2}+z^{2}\right)-\tfrac{1}{16} \tfrac{x^{2} y^{2} z^{2}}{S^{2}} . \end{gathered} \] From de Gua’s theorem, \[ \begin{aligned} & S^{2}=S_{D A B}^{2}+S_{D B C}^{2}+S_{D A C}^{2} \\ & S^{2}=\tfrac{1}{4}\left(x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2}\right) \end{aligned} \] from which

(14) \[ D O^{2}+D H^{2}=\tfrac{1}{4}\left(x^{2}+y^{2}+z^{2}\right)-\tfrac{1}{4} \tfrac{x^{2} y^{2} z^{2}}{\left(x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2}\right)} . \]

From (13) we have

(15) \[ R^{2}=\tfrac{\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)\left(x^{2}+z^{2}\right)}{4\left(x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2}\right)} . \]

After carrying out the computations in (14) and (15), we obtain \[ D O^{2}+D H^{2}=R^{2} \]

Acknowledgements

The second named author was partially supported by the Bulgarian National Science Fund, Ministry of Education and Science of Bulgaria under contract KP-06-N82/6.

REFERENCES

Coxeter, H. S. M., Greitzer, S. L. (1967). Geometry Revisited. MAA.

Johnson, R. A. (2007). Advanced Euclidean Geometry. Dover.

Prasolov, V. V. (2001). Problems in Geometry. AMS.