# Does there exists infinite point $P$ which divideds a triangle in equal area?

Given a $$\Delta ABC$$, there exists a point $$P$$ such that $$[ABP]=[ACP]=[BCP]$$ How many such a points $$P$$ exists? (Where$$[.]$$ represents area)

I previously belieced that $$P$$ is only a single point which is only median. Then after I realized that other points other than median also exists. Then I took a random point $$(x,y)$$. And three sides of triangle represented by $$(x_i,y_i)$$ but I was not able to prove. Can someone help me figure it out?

• I believe that there exists four such points.
– user1135351
Jan 8, 2023 at 10:48
• Does the point $P$ have to be within the triangle? Jan 8, 2023 at 10:50
• @SaiMehta anywhere in the plane. Jan 8, 2023 at 10:54
• @Siddharth can you prove it? Jan 8, 2023 at 10:55
• I think the only point which works inside the triangle is the centroid, $G$, which is the intersection of the medians. Jan 8, 2023 at 11:06

Let $$A=\begin{pmatrix}0\\0\end{pmatrix}$$ and $$B,C\in\mathbb R^2$$ be such that $$B$$ and $$C$$ are linearly independent. We will show that there exist exactly four points $$P\in\mathbb R^2$$ such that $$[ABP]=[APC]=[BCP]$$. In the image below, these four points are $$P,A',B',C'$$. We use the notation $$X=\begin{pmatrix}x_1\\x_2\end{pmatrix}$$ for points. We assume without loss of generality that $$b_2=0$$ and $$c_2>0$$.

First, we will show that $$[ABX]=[AXC]$$ if and only if $$X\in AD\cup AB'\subseteq\mathbb R^2$$ is in the union of the lines $$AD$$ and $$AB'$$ (not the segments!), where $$B'=C-B$$ and $$D=B+\frac 12B'=\frac 12(B+C)$$. Recall that $$[ABD]=\frac{1}{2}\det(B\; D)=\frac 12(b_1d_2-d_1b_2)=\frac 14\det(B\,C)=\frac 12[ABC]$$. Analogously, we obtain $$[ADC]=\frac 12[ABC]=[ABD]$$. The intercept theorem suggests that for any point $$X\in AD$$ the height of $$ABX$$ with respect to $$AB$$ is $$c$$ times the height of $$ABD$$ with respect to $$AB$$, where $$X=cD$$. This suggests that $$[ABX]=c[ABD]$$. The same reasoning applied to $$AXC$$ gives $$[AXC]=c[ADC]=c[ABD]=[ABX]$$. Next, notice that $$[ABB']=\frac 12\det(B\,B')=\frac 12\det(B\,C)=[ABC]$$ and similarly $$[ACB']=[ABC]=[ABB']$$. Another application of the intercept theorem gives $$[AXC]=[ABX]$$ for all $$X\in AB'$$, so indeed we have $$[ABX]=[AXC]$$ for all $$X\in AD\cup AB'$$.

Now we're left to show that the other direction holds. For this purpose let $$X\in\mathbb R^2\setminus\{A\}$$ be such that $$[ABX]=[AXC]$$, let $$c=\|B\|_2$$, $$b=\|C\|_2$$ and $$x=\|X\|_2$$. Further, let $$\alpha=\angle BAC$$ and $$\omega=\angle BAX$$. Using the intercept theorem as before, we notice that $$[ABY]=[AYC]$$ for all $$Y\in AX$$, so we assume without loss of generality that $$x_2\ge 0$$, so we have $$0\le\alpha,\omega<\pi$$. Recall that we have $$[ABX]=\frac 12x|\sin\omega|c$$ and similarly $$\frac 12x|\sin(\alpha-\omega)|b=[AXC]=[ABX]=\frac 12x|\sin\omega|c$$, which gives $$|\sin(\omega-\alpha)|=\frac cb\sin\omega$$. Using $$\alpha>0$$, since $$c_2>0$$, yields that this equation has exactly two solutions on $$[0,\pi]$$. Hence, we have exactly four solutions for arbitrary $$X$$.

At this point, we have identified all $$X$$ for which $$[ABX]=[AXC]$$. By symmetry of the problem we have $$[ABX]=[BCX]$$ exactly for $$X\in BE\cup BC'$$ where $$E=\frac 12C$$ and $$C'=B-C$$, and $$[AXC]=[BCX]$$ exactly for $$X\in CF\cup CA'$$, where $$F=\frac 12B$$ and $$A'=B+C$$.

It's left to show that the intersection of all of these three sets is $$\{P,A',B',C'\}$$, where $$P=\frac 13(A+B+C)$$. Notice that $$a'_2=c_2=b'_2>0$$, that $$b'_1=c_1-b_1, and that $$c'_2=-c_2<0$$, so $$A'B'C'$$ is a proper triangle. Computing the intersection points of the lines is mechanical (e.g. $$A'$$ lies on $$BA'$$ by definition, on $$AD$$ since $$A'=2D$$ and on $$CA'$$ by definition, and these lines are clearly distinct, so it is the only intersection point of each pair). This completes the proof.

Case 1: $$P$$ lies outside the triangle

WLOG let $$P$$ be on the opposite side of $$AB$$ to $$C$$, so that $$[ABC]+[ABP]=[ACP]+[BCP]$$Since $$[ABP]=[ACP]=[BCP]$$, as mentioned in the question, $$[ABC]=[ABP]=[ACP]=[BCP]$$. Since $$[ACP]=[BCP]$$, $$CP$$ is the perpendicular bisector of $$AB$$. Since $$[ABC]=[ABP]$$, $$AB$$ is the perpendicular bisector of $$CP$$. Therefore, quadrilateral $$ACBP$$ is a rhombus and $$P=A+B-C$$. Doing this where $$P$$ is on the opposite side of the other two edges also gives $$P=B+C-A$$ and $$P=C+A-B$$ as solutions.

Case 2: $$P$$ lies inside the triangle

Let $$D$$, $$E$$ and $$F$$ be the midpoints of $$BC$$, $$CA$$ and $$AB$$ respectively. Since using the area of a triangle as $$\frac{1}{2}bh$$, we can see that $$[AEP]=[AFP]=[BFP]=[BDP]=[CDP]=[CEP]$$, as each midpoint bisects their respective edges. We can see that $$P$$ is the intersections of the medians of triangle $$ABC$$, and so is the centroid, so $$P=\frac{1}{3}(A+B+C)$$.

• CP is perpendicular bisector of AB is not necessarily true in your 1st case. It's only bisector
– user1135351
Jan 8, 2023 at 12:04
• Yes, that is true. Would it be right to say that the perpendicular distance from $C$ to $AB$ is the same as the perpendicular distance from $P$ to $AB$? Jan 8, 2023 at 12:06
• Yes, that's absolutely true. You can prove it by congruence.
– user1135351
Jan 8, 2023 at 12:08
• Also $ACBP$ isn't a rhombus, you can call it a parallelogram.
– user1135351
Jan 8, 2023 at 12:09

First of all make a triangle $$\Delta ABC$$ draw medians $$AE,BD,CF$$ and call $$AE∩BD∩CF≡G$$ (which is also the median)

I am proving with respect to one side, similarly you can prove for others.

If point is inside it will be median, I'm not proving for median, but for the point which contained in $$∠ABC$$.

Let's start,

Extended $$BD$$ to $$B'$$ such that $$BD=B'D$$, you can see in the diagram above.

Let, $$∠CDB'=∠ADG=\alpha$$.

Note the altitude of $$\Delta ABB'$$ and $$\Delta CBB'$$ to $$BB'$$ are equal.

Hint: prove it using $$AD=DC$$ and $$\alpha$$.

So, $$[ABB']=[CBB']$$

Now to prove that, $$[ACB']=[ABB']⇒[CDB']=[ADB]$$

Note that, $$[CDB']= \frac{1}{2}CD.DB'.\sin\alpha$$ But $$CD=AD, DB'=BD$$. So, $$[CDB']= \frac{1}{2}AD.BD.\sin\alpha=[ABD]$$ Hence, $$[ABB']=[ACB']=[CBB']$$

So, $$B'$$ is one of the point you were looking for, and only one point which is contained by $$∠ABC$$. Similarly $$C',A'$$

Hence there are four such points which here are $$A',B',C',G$$

• Can you please improve the readability? There seem to be several, at least formal, mistakes. Also, it seems that you have only shown the existence of at least four points, so one direction is still missing. Jan 8, 2023 at 17:23
• The user asked just the existance. Which I had proved. He doesn't asked to locate the point or something else.
– user1135351
Jan 8, 2023 at 18:19

For interior point $$P$$: We recall the classical proof of Ceva's theorem. Let $$AX$$, $$BY$$, $$CZ$$ be the Cevians through $$P$$. We have $$\frac{BX}{XC}=\frac{(ABX)}{(ACX)}=\frac{(PBX)}{(PCX)}=\frac{(ABX)-PBX}{(ACX)-PCX}=\frac{(ABP)}{(ACP)}=1$$ So, $$BX=XC$$. We similarly show that $$CY=YA$$ and $$AZ=ZB$$ so $$P=G$$ is the centroid.

For exterior point $$P_a$$ which is on the side determined by the interior angle $$\angle A$$: Let $$D=AP_a\cap BC$$ and areas $$(BP_aD)=x$$ and $$(P_aCD)=y$$. Then $$(ABD)=y$$ and $$(ACD)=x$$. Dropping perpendiculars $$BH_1$$ and $$CH_2$$ from $$B$$ and $$C$$ on the diagonal $$AP_a$$ respectively, we have equivalent right triangles $$\triangle BH_1D$$ and $$\triangle CH_2D$$ due to the fact that $$(ABP_a)=(ACP_a)=x+y$$ implies $$BH_1=CH_2$$. Hence, $$BD=DC$$, $$AD$$ is median and $$x=y$$. Summary: $$P_a$$ is the unique point such that $$\square ABP_aC$$ is a parellogram.

Similarly, we have another two such exterior points $$P_b$$ and $$P_c$$ on the sides determined by the interior angles $$\angle B$$ and $$\angle C$$ respectively. Total we have 4 points with required property.