Good solution . But I will propose another one. I used the Menelaus Theorem considering triangle ADC with the line BME. I arrived with a relation for (MD/MA)=(4/5) I draw the line CM and I named the area of [BMD]=S. Then I worked backwards to find [MDC]=(2/3)*S Now i used the area relation for triangles AME with EMC which is (3/4). Area relation between triangles AMC with CMD, which is (5/4). Then I got area [AME]=(15/42)*S and area [EMC]=(20/42)*S Area relation between triangles ABM and BMD, which gave me area of [ABM]=(5/4)*S Finally added up to 1 all bits and pieces of relative areas. This gave me area [BMD]=S=(4/15)
For any arbitrary pair of triangles with unit area, we can find an area-preserving affine transformation that maps one to the other, so w.l.o.g. assume that ABC is a right triangle with right angle at the origin and sides CB=1 and CA=2 along the positive x- and y-axes, respectively. In homogeneous coordinates, using the intercept-intercept form, we find that the line BE is [1,1/(8/7),-1], AD is [1/(2/5),1/2,-1], and the coordinates of point M are their cross product. The height of the blue triangle is just the y-coordinate of M. Extract it, multiply by the base (3/5) and divide by 2.
Solving such a problem could be achieved in more direct approach by relating the areas of the triangles to each other and then setting a linear system of equations in three unknown one of them the required area. Solving these equations for [MBD] we find it equal to 4/15. However, Menelaus's theorem application to the problem is the more elegant method to solve it.
THANKS, Great exercise as always. We know that the barycentric coordinates (en.wikipedia.org/wiki/Barycentric_coordinate_system) of a point M in the affine frame (A,B,C) are (areaMBC,areaAMC,areaABM)/areaABC. From what I understand from what @femalesworld2 says, the barycentric coordinates of M in the affine frame (A,B,C) are (4,2,3)/(4+2+3) =(4/9,2/9,3/9). We deduce that areaMBC/areaABC=4/9 and therefore that areaMBC=4/9 areaABC but as areaBMD=3/5 areaBMC, we deduce that areaBMD=3/5*4/9*areaABC=4/15.
Let AC be considered the base of ΔABC. We note that ΔBCE and ΔABC have the same height but the base of ΔBCE is 4/7 the length of the base of ΔABC. Thus, ΔBCE has 4/7 the area of ΔABC. Area ΔABC equals 1, so area ΔBCE equals 4/7. We construct EN as in the video and determine that, if CD has length 2a, length DN = 6a/7 and length CN = 8a/7. To simplify the diagram, we can now discard line segments AB, AM and AE. Consider BC to be the base of ΔBCE. Then, ΔBEN and ΔBCE have the same height. The portion of ΔBCE's area which belongs to ΔBEN is the ratio of their bases, BN/BC = (3a + 6a/7)/5a = (27a/7)/5a = 27/35, so ΔBEN has area (27/35)(4/7) = 108/245. Now we note that ΔBMD and ΔBEN are similar. We find the ratio of corresponding sides BD/BN = 3a/(3a + 6a/7) = 3a/(27a/7) = 21/27 = 7/9. Area of ΔBMD is the square of the corresponding sides ratio times the area of ΔBEN. So, area ΔBMD = (7/9)²(108/245) = (49)(108)/(81)(245) = (49)(27)(4)/(27)(3)(49)(5) = 4/(3)(5) = 4/15, as Math Booster also found.
We can make unlimited different triangles of [ABC]=1, and all the four small triangles area will change with new ABC, so it needs more restrict info to get limited answers.
My solution: (In all calculations below, when I use ratios of areas of triangles, these have the same height, so the ratio of their areas is equal to the ratios of their bases). Draw line CM [BMD] = x → [CMD]/[BMD] = 2a/(3a) = 2/3 → [CMD] = 2/3 x [AME] = y → [CME]/[AME] = 4b/(3b) = 4/3 → [CME] = 4/3 y [ABE]/[CBE] = 3b/(4b) = 3/4 → [ABE] = 3/4 [CBE] → [AMB] + [AME] = 3/4 ([BMD] + [CMD] + [CME]) → [AMB] + y = 3/4 (x + 2/3 x + 4/3 y) = 5/4 x + y → [AMB] = 5/4 x AM/DM = [AMB]/[BMD] = (5/4 x) / x = 5/4 → AM = 5c, DM = 4c [AMC]/[DMC] = 5c/(4c) = 5/4 → [AMC] = 5/4 [DMC] = 5/4 (2/3 x) = 5/6 x [ABC] = 1 → [BMD] + [CMD] + [AMC] + [AMB] = 1 → x + 2/3 x + 5/6 x + 5/4 x = 1 → 15/4 x = 1 → x = [BMD] = 4/15
If MC is connected then also by the triangle area law MBD /MDC = 3/2 Cme / ema=4/3 Abd / adc =3/2 Now by manipulating these equations bmd can also be found as 4/15
I divided the triangle into 4 parts as ABM, AEM, CDEM, BDM. The areas are 10/35, 5/35, 9/35, 11/35 respectively. So my value for that area is 11/35. As you see, there is more than one solution to this problem, and the values work for the side length segments of the triangle. Because there are 3 equations and 4 variables, therefore, there is no unique solution. The figure is not verified.
The trick is to further divide CDME into 2 triangles: CDM, CEM, then you can make additional equations for these 2 areas. This will allow you solve for a unique solution for the areas of all 5 triangles.
I don't see anything geometric, so solved by brute force equations ... (sorry I love maths but I hate geometry) SO ... lets def coord , for ex. D(0,0) B(-3a,0) C(2a,0) E(2a-4bm,4bn) A(2a-7bm,7bn) m²+n²=1 (will not be useful ...) and -n/m is slope of AC M(x,y) DMA is a line : x/y = (2a-7bm)/7bn BME is a line : (x+3a)/y = (5a-4bm)/4bn x/y + 3a/y = (2a-7bm)/7bn + 3a/y = (5a-4bm)/4bn 3a/y = (5a-4bm)/4bn - (2a-7bm)/7bn = (35a-28bm-8a+28bm)/28bn = 27a/28bn y/3a = 28bn/27a y = 28bn/9 Area BMD = [1/2].3a.28bn/9 = 14abn/3 Area ABC = [1/2].5a.7bn = 35abn/2 = 1 abn = 2/35 Area BMD = (14.2)/(3.35) = 4/15 simplifies by all parameters !! It's long but I'm more confident in pure equations.
![Уравнение, которое меняет взгляд на мир [Veritasium]](http://i.ytimg.com/vi/DH1cv0Rdf2w/mqdefault.jpg)
Good solution . But I will propose another one.
I used the Menelaus Theorem considering triangle ADC with the line BME. I arrived with a relation for (MD/MA)=(4/5)
I draw the line CM and I named the area of [BMD]=S. Then I worked backwards to find [MDC]=(2/3)*S
Now i used the area relation for triangles AME with EMC which is (3/4).
Area relation between triangles AMC with CMD, which is (5/4). Then I got area [AME]=(15/42)*S and area [EMC]=(20/42)*S
Area relation between triangles ABM and BMD, which gave me area of [ABM]=(5/4)*S
Finally added up to 1 all bits and pieces of relative areas. This gave me area [BMD]=S=(4/15)
For any arbitrary pair of triangles with unit area, we can find an area-preserving affine transformation that maps one to the other, so w.l.o.g. assume that ABC is a right triangle with right angle at the origin and sides CB=1 and CA=2 along the positive x- and y-axes, respectively. In homogeneous coordinates, using the intercept-intercept form, we find that the line BE is [1,1/(8/7),-1], AD is [1/(2/5),1/2,-1], and the coordinates of point M are their cross product. The height of the blue triangle is just the y-coordinate of M. Extract it, multiply by the base (3/5) and divide by 2.
Solving such a problem could be achieved in more direct approach by relating the areas of the triangles to each other and then setting a linear system of equations in three unknown one of them the required area. Solving these equations for [MBD] we find it equal to 4/15. However, Menelaus's theorem application to the problem is the more elegant method to solve it.
THANKS, Great exercise as always. We know that the barycentric coordinates (en.wikipedia.org/wiki/Barycentric_coordinate_system) of a point M in the affine frame (A,B,C) are (areaMBC,areaAMC,areaABM)/areaABC. From what I understand from what @femalesworld2 says, the barycentric coordinates of M in the affine frame (A,B,C) are (4,2,3)/(4+2+3) =(4/9,2/9,3/9). We deduce that areaMBC/areaABC=4/9 and therefore that areaMBC=4/9 areaABC but as areaBMD=3/5 areaBMC, we deduce that areaBMD=3/5*4/9*areaABC=4/15.
Let AC be considered the base of ΔABC. We note that ΔBCE and ΔABC have the same height but the base of ΔBCE is 4/7 the length of the base of ΔABC. Thus, ΔBCE has 4/7 the area of ΔABC. Area ΔABC equals 1, so area ΔBCE equals 4/7. We construct EN as in the video and determine that, if CD has length 2a, length DN = 6a/7 and length CN = 8a/7. To simplify the diagram, we can now discard line segments AB, AM and AE. Consider BC to be the base of ΔBCE. Then, ΔBEN and ΔBCE have the same height. The portion of ΔBCE's area which belongs to ΔBEN is the ratio of their bases, BN/BC = (3a + 6a/7)/5a = (27a/7)/5a = 27/35, so ΔBEN has area (27/35)(4/7) = 108/245. Now we note that ΔBMD and ΔBEN are similar. We find the ratio of corresponding sides BD/BN = 3a/(3a + 6a/7) = 3a/(27a/7) = 21/27 = 7/9. Area of ΔBMD is the square of the corresponding sides ratio times the area of ΔBEN. So, area ΔBMD = (7/9)²(108/245) = (49)(108)/(81)(245) = (49)(27)(4)/(27)(3)(49)(5) = 4/(3)(5) = 4/15, as Math Booster also found.
We can make unlimited different triangles of [ABC]=1, and all the four small triangles area will change with new ABC, so it needs more restrict info to get limited answers.
My solution:
(In all calculations below, when I use ratios of areas of triangles, these have the same height, so the ratio of their areas is equal to the ratios of their bases).
Draw line CM
[BMD] = x → [CMD]/[BMD] = 2a/(3a) = 2/3 → [CMD] = 2/3 x
[AME] = y → [CME]/[AME] = 4b/(3b) = 4/3 → [CME] = 4/3 y
[ABE]/[CBE] = 3b/(4b) = 3/4
→ [ABE] = 3/4 [CBE]
→ [AMB] + [AME] = 3/4 ([BMD] + [CMD] + [CME])
→ [AMB] + y = 3/4 (x + 2/3 x + 4/3 y) = 5/4 x + y
→ [AMB] = 5/4 x
AM/DM = [AMB]/[BMD] = (5/4 x) / x = 5/4 → AM = 5c, DM = 4c
[AMC]/[DMC] = 5c/(4c) = 5/4 → [AMC] = 5/4 [DMC] = 5/4 (2/3 x) = 5/6 x
[ABC] = 1
→ [BMD] + [CMD] + [AMC] + [AMB] = 1
→ x + 2/3 x + 5/6 x + 5/4 x = 1
→ 15/4 x = 1
→ x = [BMD] = 4/15
If MC is connected then also by the triangle area law
MBD /MDC = 3/2
Cme / ema=4/3
Abd / adc =3/2
Now by manipulating these equations bmd can also be found as 4/15
⭐️
I divided the triangle into 4 parts as ABM, AEM, CDEM, BDM. The areas are 10/35, 5/35, 9/35, 11/35 respectively. So my value for that area is 11/35. As you see, there is more than one solution to this problem, and the values work for the side length segments of the triangle. Because there are 3 equations and 4 variables, therefore, there is no unique solution. The figure is not verified.
The trick is to further divide CDME into 2 triangles: CDM, CEM, then you can make additional equations for these 2 areas. This will allow you solve for a unique solution for the areas of all 5 triangles.
I don't see anything geometric, so solved by brute force equations ... (sorry I love maths but I hate geometry)
SO ... lets def coord , for ex.
D(0,0)
B(-3a,0)
C(2a,0)
E(2a-4bm,4bn)
A(2a-7bm,7bn)
m²+n²=1 (will not be useful ...) and -n/m is slope of AC
M(x,y)
DMA is a line : x/y = (2a-7bm)/7bn
BME is a line : (x+3a)/y = (5a-4bm)/4bn
x/y + 3a/y = (2a-7bm)/7bn + 3a/y = (5a-4bm)/4bn
3a/y = (5a-4bm)/4bn - (2a-7bm)/7bn
= (35a-28bm-8a+28bm)/28bn = 27a/28bn
y/3a = 28bn/27a
y = 28bn/9
Area BMD = [1/2].3a.28bn/9 = 14abn/3
Area ABC = [1/2].5a.7bn = 35abn/2 = 1
abn = 2/35
Area BMD = (14.2)/(3.35) = 4/15 simplifies by all parameters !!
It's long but I'm more confident in pure equations.