Show that a point inside an acute triangle is the orthocenter

Question

Point $M$ lies inside an acute triangle $\triangle ABC$ such that the circumscribed circles of $\triangle ABM,\triangle BCM$ and $\triangle ABC$ have equal radii. Show that $M$ is the orthocenter.

I have been trying to solve the problem for an hour using the law of sines because we have just studied it, but I can't come up with anything. What about a solution using vectors?

Answer 1

These definitions are from this book: 'An introduction to the modern geometry of the triangle and the circle'; Nathan Altshiller Court.You can see these and their proof in there.

1-Definition: the midpoints of segments connecting the orthocenter and vertices of a triangle is called the Euler points of the triangle.

2-Theorem: In any triangle the mid points of sides , the foot of altitudes and Euler points are cyclic (on a circle). This is called nine point circle of a triangle.

3-Triangles (AMB), (BMC), (AMC) and(ABC) are called group of orthocenter's triangles.

4-The radius of circumscribed circles of the group of orthocenter triangle are equal to the diameter of nine point circle of triangle.

Now your statement say the radius of circumscribed circles of ABM, AMC and BMC are equal, therefore M must be orthcenter of ABC.

Answer 2

Let the centre of the circle opposite vertex $A$, $B$ and $C$ be $A_{1}$, $B_{1}$ and $C_{1}$ respectively. Draw $A_{1}B$, $A_{1}C$, $CB_{1}$, $B_{1}A$, $AC_{1}$ and $C_{1}B$.

Observe that, in quadrilateral $A_{1}BC_{1}M$, all of the sides are equal; Hence,it's a rhombus. Similarly, quadrilateral $AC_{1}MB_{1}$ and $CB_{1}MA_{1}$ are rhombi.

Now, let the circumcentre of $\triangle ABC$ be $O$. Draw $OA$, $OB$ and $OC$.

$\angle BA_{1}C=\angle BA_{1}M+\angle MA_{1}C=\angle BC_{1}M+\angle MB_{1}C=2\angle BAM+2\angle MAC=2\angle BAC$

Quadrilateral $BOCA_{1}$ is a kite since $OB=OC$ and $BA_{1}=A_{1}C$.

Also, in quadrilateral $BOCA_{1}$, $\angle BOC=2\angle BAC=\angle BA_{1}C$. $\Rightarrow$ Quadrilateral $BOCA_{1}$ is a rhombus.

Similarly, it can be proven that quadrilateral $OBC_{1}A$ and $AOCB_{1}$ are rhombi as well.

$\Rightarrow OB=A_{1}C$

Hence, the circumradii of $\triangle AMB$, $\triangle BMC$ and $\triangle CMA$ are all equal to the circumradius of $\triangle ABC$.

Notice that, in $\triangle AC_{1}B_{1}$ and $\triangle OBC$, $AC_{1}=OB$, $AB_{1}=OC$ and $\angle C_{1}AB_{1}=\angle BOC$ [Since $OB\parallel AC_{1}$ & $OC\parallel AB_{1}$]

Hence, $\triangle AC_{1}B_{1}\cong \triangle ABC$ by $S-A-S$ criterion of congruence. $\Rightarrow B_{1}C_{1}=BC$

Observe that, in quadrilateral $B_{1}C_{1}BC$, $BC_{1}=B_{1}C$ and $B_{1}C_{1}=BC$. Hence, it's a parallelogram. $\Rightarrow B_{1}C_{1}\parallel BC$

Now, since $AM\perp B_{1}C_{1}$$[$$AC_{1}MB_{1}$ is a rhombus$]$ and $B_{1}C_{1}\parallel BC$, $AM\perp BC$.

Similarly,it can be proven that $BM\perp CA$. Therefore, $M$ is the orthocentre of $\triangle ABC$.

Answer 3

$\underline{\mathrm{Introduction}}$

As shown in $\mathrm{Fig.\space 1}$, we take an arbitrary point $M$ inside the acute triangle $ABC$ and draw the three cevians $AD$, $BE$, and $CF$ through it to form three subtriangles $MAB$, $MBC$, and $MCA$. After constructing the respective circumcircles of those subtriangles, we assert that two of these circles, namely $MAB$, and $MBC$, have the same radius as that of the triangle $ABC$. Please note that $A_1$, $B_1$, $C_1$, and $O$ are the centers of the mentioned circumcircles.

Our proof consists of two parts. In the first part we show that the radius of the circumcircle of the subtriangle $MCA$ is also equal to that of the triangle $ABC$. In part 2 we prove that, if the radii of all triangles $MAB$, $MBC$, $MCA$, and $ABC$ are equal, then $M$ is indeed the orthocenter of the latter. The proof is lengthy, but not difficult. In fact, this is an extended exercise of angle chasing mostly using the properties of cyclic quadrilaterals and subtended angles at centers and circumferences of circles.

$\underline{\mathrm{Part\space 1}}$

For brevity, let $\measuredangle CAB= \alpha$ and $\measuredangle DAB = \phi$. Also let the radii of the circumcircles $MCA$ and $ABC$ to be equal to $r$ and $R$ respectively. As shown in $\mathrm{Fig.\space 1}$, we draw nine auxiliary line segments $A_1M$, $A_1C$, $B_1C$, $B_1M$, $B_1A$, $C_1A$, $C_1M$, $OA$, and $OC$,

The chord $MB$ of the circle $MAB$ subtends $\measuredangle MAB$ and $\measuredangle MA_1B$ at the circumference and the center of the said circle respectively. Therefore, we have $$\measuredangle MC_1B = 2\measuredangle MAB = 2\phi.$$

Since each side that forms the quadrilateral $MC_1BA_1$ is a radius of the circumcircle of either $\triangle MAB$ or $\triangle MBC$, we can state that it is a rhombus. Because the opposite angles of a rhombus are equal, we shall write, $$\measuredangle BA_1M = \measuredangle MC_1B = 2\phi. \tag{1}$$

The chord $BC$ of the circle $ABC$ subtends $\measuredangle CAB$ and $\measuredangle COB$ at the circumference and the center of the said circle respectively. Therefore, we have $$\measuredangle COB = 2\measuredangle CAB = 2\alpha.$$

Since each side of the quadrilateral $A_1COB$ is a radius of the circumcircle of either $\triangle ABC$ or $\triangle MBC$, it is a rhombus. Because of this, it is evident that, $$\measuredangle BA_1C = \measuredangle COB = 2\alpha. \tag{2}$$

Now, using (1) and (2), we can express the apex angle $\measuredangle MA_1C$ of the isosceles triangle $MA_1C$ as, $$\measuredangle MA_1C = \measuredangle BA_1C - \measuredangle BA_1M = 2\left(\alpha-\phi\right). \tag{3}$$

The quadrilateral $MA_1CB_1$ is formed by the union of two isosceles triangles $MA_1C$ and $MCB_1$, which are not congruent. Two of its sides, namely $CB_1$ and $MB_1$, are radii of the circle $MCA$ and, therefore, has the length $r$. The other two sides are radii of length $R$, which belong to the circle $MBC$. Therefore $MA_1CB_1$ is a kite.

It is easy to observe that $$\measuredangle CAM = \measuredangle CAB - \measuredangle DAB = \alpha – \phi.$$

The chord $CM$ of the circle $MAB$ subtends $\measuredangle CAM$ and $\measuredangle CB_1M$ at the circumference and the center of the said circle respectively. Therefore, we have $$\measuredangle CB_1M = 2\measuredangle CAM = 2\left(\alpha - \phi\right). \tag{4}$$

We can conclude from (3) and (4) that a pair of opposite angles of the kite $MA_1CB_1$, namely $\measuredangle CB_1M$ and $ \measuredangle MA_1C$, are equal. This means that it is a rhombus as well. Since all four sides of a rhombus are of the same length, we have, $$r = R. $$

$\underline{\mathrm{Part\space 2}}$

Let $P$ be an arbitrary point on the part of the circumference of the circle $MBC$, which does not contain $M$. Also assume that $\measuredangle AFC = \omega$.

We have already shown in $\mathrm{Part\space 1}$ that $\measuredangle BA_1C$, which is subtended by the arc $CMB$ is equal to $2\alpha$. Therefore, its namesake $\measuredangle BA_1C$ subtended by the arc $BPC$ is equal to $360^o - 2\alpha$. Note that the same arc $BPC$ subtends $\measuredangle CMB$ at $M$ on the arc $CMB$. So, we shall write $$\measuredangle CMB = \frac{1}{2}\left(360^o - 2\alpha\right) = 180^o – \alpha.$$

$$\therefore\quad \measuredangle FME = 180^o – \alpha\quad\mathrm{as\space well}.$$

Here, we have the situation, where the opposite angles of the quadrilateral $MEAF$, namely $EAF$ and $FME$, are supplementing each other. This means that the quadrilateral $AFME$ is cyclic.

Since an exterior angle of a cyclic quadrilateral is equal to the interior opposite angle, we have $$\measuredangle CEM = \measuredangle AFM = \omega. \tag{5}$$

It is up to the poster of the question to prove that the quadrilateral $CEMD$ is cyclic too using similar arguments. Start your proof by assuming $\measuredangle BCA = \gamma$ and showing that $\measuredangle AC_1B = \measuredangle BOA =2\gamma$. Next, establish that the angle subtended by the arc $AQB$, where $Q$ is an arbitrary point on the part of the circumference of the circle $MAB$ that does not contain $M$, is equal to $360^0-2\gamma$. Finally, show that $\measuredangle EMD = \measuredangle BMA = 180^0-\gamma$.

Since $CEMD$ is cyclic, we have from (5), $$\measuredangle MDC = 180^0 – \measuredangle CEM = 180^0 - \omega. \tag{6}$$

In order to conclude our proof, we need to verify yet another cyclic quadrilateral. The chord $MB$ of the circle $MBC$ subtends $\measuredangle BCM$ and $\measuredangle BA_1M$ at the circumference and the center of the said circle respectively. Therefore, we have $$\measuredangle BCM = \frac{1}{2}\measuredangle BA_1M = \phi. $$

Consequently, $\measuredangle DCF$ and $\measuredangle DAF$ of the quadrilateral $AFDC$ are equal making it cyclic. $$\therefore\quad \measuredangle ADC = \measuredangle AFC = \omega \tag{7}$$

The scenario resulted from the arguments submitted in $\mathrm{Part\space 2}$ is displayed in $\mathrm{Fig.\space 2}$.

Although the two angles $\measuredangle ADC$ and $\measuredangle MDC$ are one and the same, it can be expressed in two different ways according to (6) and (7). Therefore, $$\omega = 180^0 - \omega \quad\rightarrow\quad \omega = 90^0.$$

This proves that each cevian is perpendicular to the side opposite to the vertex, from which it is dropped. In other words, the three cevians are the altitudes of $\triangle ABC$. Since altitudes are concurrent at the orthocenter of the triangle, $M$ is the orthocenter of $\triangle ABC$.

Answer 4

Here is a detailed solution using inversion. In the conditions from the OP let $D$, $E$, $F$ be the feet of the heights from $A$, $B$, $C$ respectively.

The points $A,C,D,F$ are on a circle, so let us consider the inversion denoted by $\mathcal I=*$, so $X\to \mathcal I(X)=X^*$,

• with center in $B$ and
• with power $k^2:=BF\cdot BA=BD\cdot BC=ac\cos B$.

So $A^*=F$, $F=A^*$; $C^*=D$, $D^*=C$. In a picture:

Let $O$ be the circumcenter of $\Delta ABC$. Let $B_1\in BO$ be opposed to $B$ on the circumcircle $\odot(O)=\odot(ABC)$, so $BB_1=2BO=2R$. We transform it into the point $B_1^*\in BO$. Now we have the following simple observations that will restrict the position of the point $M$ from the OP, not mentioned on the picture, but mentioned below.

• The circle $\odot(ABC)=\odot(O)$ is transformed into the line $A^*C^*=DF$. It contains the point $B_1^*$, which realizes the minimal distance from $B$ to a point on $A^*C^*=DF$, since $B_1$ realizes the maximal distance from $B$ to a point of the circle $\odot O$. So $BB_1^*\perp DF$.
• The circle $\odot(BMA)$ maps via $\mathcal I$ into the line $M^*A^*=M^*F$, and since this circle is obtained by a rotation around $B$ of $\odot(O)$, (same radius,) its $\mathcal I$-image is also a corresponding rotation of $DF=\odot(O)^*$, in particular also tangent to the circle $$(\beta)$$ centered in $B$ with radius $BB_1^*$. So $M^*$ is on the tangent from $F$ to $(\beta)$.
• Similarly, $M^*$ is on the tangent from $D$ to $(\beta)$.

It remains to show that these two tangents intersect in the point $E$. (From here, $M^*$ is determined to be $E$, thus $M=M^{**}=E^*=H$, the orthocenter of the given triangle. We use $EHDC$ cyclic, thus $BE\cdot BH= BD\cdot BC=k^2$, thus $E$ transforms to $H$.)

(The reader may want to stop the lecture here, and complete the proof on her or his own.)

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Let us show that $EF$ is tangent $(\beta)$. By the same inversion (which conserves orthogonality), we show equivalently that the circle $\odot(BHA)=\odot(BE^*F^*)=(EF)^*$ is tangent to $(\beta)^*$, the circle centered in $B$ with radius $BB_1=2BO=2R$.

For this, let $B_1'\in(\beta)^*$ be the reflection of $B_1$ in $AB$. Since

• $\Delta B_1'AB=\Delta B_1AB$ have each a right angle in $A$, so $BB_1$ and $BB_1'$ are respectively diameters for their circumcircles,
• and $\odot(ABB_1')$ (with diameter $BB_1'$) is tangent in $B_1'$ to $(\beta)^*$ (with center $B$ and radius $BB_1'=BB_1=2R$)

it is enough to show that $H$ is also on $\odot(AB_1'B)$. This is immediate: $$ \widehat{AB_1'B} = \widehat{AB_1B} = \widehat{ACB} = \widehat{ECB} = \widehat{DHB} = 180^\circ -\widehat{AHB} \ . $$ $\square$