If you put C(0,0) and A(10,0) and B is in the x>0 , y>0 area, CF is the locus of all points where a square is in the triangle, one side is on AC and another vertex is on CB... so you can see that if CB is y=mx then CF is the y= [m/(m+1)]x and you can find F . [ locus of all points of the forth vertex]
i have an idea: vary xf until the perpendicular distance to the left hand line is equal to the intersection on the right hand line while yf=0 all the time 10 print "premath-olympiad mathematics-find area of the pink square":dim x(4,2),y(4,2) 20 la=10:lb=10:lc=12:sw=la/(la+lb+lc):lh=(la^2-lb^2+lc^2)/2/lc:h=sqr(la^2-lh^2) 30 xp=sw:yp=0:yf=yp:x1=0:y1=0:x2=lh:y2=h:xg21=lc:yg21=0:xg22=lh:yg22=h:yg11=0 40 dx=x2-x1:dy=y2-y1:xa=0:ya=0:xc=lh:yc=h:goto 180 50 zx=dx*(xp-x1):zy=dy*(yp-y1):k=(zx+zy)/(dx^2+dy^2) 60 dxk=dx*k:dyk=dy*k:xe=x1+dxk:ye=y1+dyk:xf=xp:dxu=xe-xf:dyu=ye-yf:n=sqr(dxu^2+dyu^2) 70 xg11=xf:xg12=xg11+dyu:yg12=yg11-dxu 80 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12) 90 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22) 100 a13=a131+a132:a23=a231+a232:gosub 110:goto 160 110 ngl1=a12*a21:ngl2=a22*a11 120 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end 130 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2 140 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2 150 xl=zx/ngl:yl=zy/ngl: return 160 ls=sqr((xf-xe)^2+(yf-ye)^2):lr=sqr((xf-xl)^2+(yf-yl)^2) 170 dg=(ls-lr)/lh:return 180 gosub 50 190 xp1=xp:dg1=dg:xp=xp+sw:xp2=xp:gosub 50:if dg1*dg>0 then 190 200 xp=(xp1+xp2)/2:gosub 50:if dg1*dg>0 then xp1=xp else xp2=xp 210 if abs(dg)>1E-10 then 200 else print "die flaeche des quadrats=";ls^2 220 lae=sqr((xp-xa)^2-ls^2):dx=lh:dy=h:n=sqr(dx^2+dy^2):dxd=(xc-xa)*(lae+ls)/la 230 xd=xa+dxd:dyd=(yc-ya)*(lae+ls)/la:yd=ya+dyd 240 masy=900/h:masx=1200/lc:if masx run in bbc basic sdl and hit ctrl tab to copy from the results window
ABC and FGB are similar having angle in B in common and CAB = GFB because AB and FG are parallel. Split triangle GFB in two right triangle with height GH. Their sides are GH=8a, BH=6a and GB=10a GHB and AEF are similar being right and because angles in B and A are congruent being ABC isosceles, so GH : EF = GB : AF 8a : 10a = 10a : AF AF = 25/2a AB = AF + FH + BH = 12 AB = 25/2a + 6a + 6a = 12 a = 24/49 side of square = 10a => s = 10*24/49 = 240/49
Very smart solution! 🤣 I took a quick look on the arbitrary solution for the isoceles triangle. Let the base of triangle be *c* and the congruent legs be *a*. Then, height h will be h = √(a²-c²/4). After a little math mumbo-jumbo one will get: x(a,c) = [a·c·√(4a²-c²)] / [2a² + c·√(4a²-c²)], a rather complicated term. In this case, we have c = 12, a = 10, therefore x = 10·12·√(4·10²-12²) / [2·10² + 12·√(4·10²-12²)], which gives x = 240/49 ≈ 4.89796. x² = 23.99... *Challenge:* Where are the vertices E, D of the square lying on the lefty a-side?
I thought because the triangles ABC and BFG are similar triangles, you could use the angle-angle theorem angle at point B = alpha angle at point C = angle at point G = beta, because both are isosceles triangles AC/AB = FG/BF 10/12 = FG/6 --> FG = 5 area square = FG² = 25 Also 12a = 6, so a = 0.5. side square = 10a = 10*0.5 = 5 What have I done wrong here ?
At a first glimpse I would say that the Area of the Pink Square is ~ 5^2 su or ~ 25 su But if the Point F is the Middle Point between A and B, my answer is 4,8^2 su ~ 23,04 su. I'll be back later to prove it.
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23.999 or 24 n=4.89796, the length of the square let the side of the square = n the area of triangle CDG + AEF = (10-n)(n)/2 Since both have the same base 'n' and 10-n is the difference of the side n of the square the area of CDG + AEF = (10 n- n^2)/2 one more triangle to go: BFG Since BFG is similar to ABC, and CF is 8 (3-4-5 triangle), then the two legs of BFG in terms of 'n' are 8/10 (n) and 12/10 (n) Hence, the BFG area in terms of 'n' = [ (0.8n)(1.2n)]/2 = 0.96n^2/2 Area of ABC = 12 * 9 /2 = 48, then this equal the area of CDB + AEF, + area of the square + are of BFG = 0.96n/2 Hence 48 = (10n-n^2)/2 + n^2 + 0.96n^2/2 96 = 10 n- n^2 + 2n^2 + 0.96n^2 (multiply both sides by 2) 0 = 1.96n^2 + 10n - 96 n= 4.89796 (Quadratic formula calculator so the length of the side of the square = n I had first put the side of the square, but you want the area area =4.89796 x 4.89796 = 23.9999
@@skwest Thanks for this comment, but I meant that the C and "X' (the midpoint of AB ) is 8 before I looked at the Premath labeling of the diagram: ABCDEFG after working the problem on a sheet of paper. So, when I did look, I somehow used "F" for "X." Thanks again, as you are correct as though DE lies on AF , DG does not lie on CB.
I, apparently, am the most pig-headed algebra guy, ever. I found this problem to be intensely annoying. Approached it 3× from the beginning, because I just couldn't do the algebra without being confused. In the end though, several intersecting equations-of-lines solved this. First, it is critical to recognize that the height of the ABC triangle is 8 units. Because … ½ of 12 base units with a hypotenuse of [10] units leaves [8] units as the rise. Pythagoras. This in turn defines f(𝒙) = ⁸⁄₆𝒙 ⊕ 0 f(𝒙) = ⁴⁄₃𝒙 ⊕ 0 [⊕0] because it has a (0, 0) origin. Right? Then the diagonal “side of the square” (lower one, left) has an inverted line equation: f'(𝒙) = -¾𝒙 + B So what is the B? Ah … well introduce a new term [𝒂], the bit along the base, from the left which intersects with the corner of where the square meets the rising left triangle side. Because of the line function (the f(𝒙) one), we know that the height is 4𝒂/3 so -3𝒂/4 + B = 4𝒂/3 … shift around B = 4𝒂/3 ⊕ 3𝒂/4 … find a common denominator and combine B = 25𝒂/12 So, that can be put into the inverted line equation f'(𝒙) = -¾𝒙 + 25𝒂/12 We can use this immediately to find the point on the base, where it intersects. 0 = -¾𝒙 + 25𝒂/12 … rearrange to solve for 𝒙 𝒙 = 25𝒂/9 ________________________________________ In a similar line of reasoning, the diagonal of the square that rises from the base to the right side is h(𝒙) = 4𝒙/3 + B = 0 … at 𝒙 = 25𝒂/9 so 0 = ⁴⁄₃ × 25𝒂/9 + B B = -100𝒂/27 … so h(𝒙) = 4𝒙/3 - 100𝒂/27 𝒔² = 400𝒂²/81 … thru lots of little algebra With that we can find where it intersects at g(𝒙), the equation for the right side of the large containing triangle. g(𝒙) = -4𝒙/3 + 16 (because must be 0 at [𝒙 = 12]) Working through a lot of algebra setting [h(𝒙) = g(𝒙)], Icame up with 𝒙 = 6 + 75𝒂/54 Since this lower right triangle is SIMILAR to the △ABC overall one, which is isosceles, we know that 𝒔² = (6 - 75𝒂/54)² + (⁴⁄₃)²(6 - 75𝒂/54)² … reduces to 𝒔² = ²⁵⁄₉(6 - 75𝒂/54)² … thus also 400𝒂²/81 = ²⁵⁄₉(6 - 75𝒂/54)² … which expands to 441𝒂² - 48600𝒂 + 104976 = 0 … being quadratic has roots 𝒂 = [ 108 or 2.2041 ] Clearly it cannot be the larger one, so 2.2041 ought to work. 𝒔² = 400 × 2.2041² ÷ 81 𝒔² = 23.9900 Or about 24. Yay! HARD work. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
Essendo un triangolo isoscele conosco gli angoli e,con il teorema dei seni calcolo a=300/49,b=288/49,a+b=12..per cui l=300/49sin(arccos6/10)=300/49*4/5=240/49
I used Sin α = 0.8 to solve the side of the square EF = x. CF / AC = EF / AF 8 / 10 = X / 6 X = 4.8 AREA = 4.8^2 = 23.04 What is wrong with my approach?
This topic is for advanced students so please avoid over explaining how to add the fractions. Just give the answer straight away. That will make the content more relevant
Very nice and clear explanation. Well done.
Thanks❤️🌹
Brilliant!
Draw perpendicular to AC from B, intersecting AC at P. BP=9.6, DG=x, CG=10-x, DG:BP=CG:CB, x:9.6=(10-x):10, x=240/49
If you put C(0,0) and A(10,0) and B is in the x>0 , y>0 area, CF is the locus of all points where a square is in the triangle, one side is on AC and another vertex is on CB... so you can see that if CB is y=mx then CF is the y= [m/(m+1)]x and you can find F . [ locus of all points of the forth vertex]
Merry Christmas Premath and all the subscribers
i have an idea: vary xf until the perpendicular distance to the left hand line is equal to the intersection on the right hand line while yf=0 all the time
10 print "premath-olympiad mathematics-find area of the pink square":dim x(4,2),y(4,2)
20 la=10:lb=10:lc=12:sw=la/(la+lb+lc):lh=(la^2-lb^2+lc^2)/2/lc:h=sqr(la^2-lh^2)
30 xp=sw:yp=0:yf=yp:x1=0:y1=0:x2=lh:y2=h:xg21=lc:yg21=0:xg22=lh:yg22=h:yg11=0
40 dx=x2-x1:dy=y2-y1:xa=0:ya=0:xc=lh:yc=h:goto 180
50 zx=dx*(xp-x1):zy=dy*(yp-y1):k=(zx+zy)/(dx^2+dy^2)
60 dxk=dx*k:dyk=dy*k:xe=x1+dxk:ye=y1+dyk:xf=xp:dxu=xe-xf:dyu=ye-yf:n=sqr(dxu^2+dyu^2)
70 xg11=xf:xg12=xg11+dyu:yg12=yg11-dxu
80 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12)
90 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22)
100 a13=a131+a132:a23=a231+a232:gosub 110:goto 160
110 ngl1=a12*a21:ngl2=a22*a11
120 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end
130 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2
140 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2
150 xl=zx/ngl:yl=zy/ngl: return
160 ls=sqr((xf-xe)^2+(yf-ye)^2):lr=sqr((xf-xl)^2+(yf-yl)^2)
170 dg=(ls-lr)/lh:return
180 gosub 50
190 xp1=xp:dg1=dg:xp=xp+sw:xp2=xp:gosub 50:if dg1*dg>0 then 190
200 xp=(xp1+xp2)/2:gosub 50:if dg1*dg>0 then xp1=xp else xp2=xp
210 if abs(dg)>1E-10 then 200 else print "die flaeche des quadrats=";ls^2
220 lae=sqr((xp-xa)^2-ls^2):dx=lh:dy=h:n=sqr(dx^2+dy^2):dxd=(xc-xa)*(lae+ls)/la
230 xd=xa+dxd:dyd=(yc-ya)*(lae+ls)/la:yd=ya+dyd
240 masy=900/h:masx=1200/lc:if masx
run in bbc basic sdl and hit ctrl tab to copy from the results window
ABC and FGB are similar having angle in B in common and CAB = GFB because AB and FG are parallel.
Split triangle GFB in two right triangle with height GH. Their sides are
GH=8a, BH=6a and GB=10a
GHB and AEF are similar being right and because angles in B and A are congruent being ABC isosceles, so
GH : EF = GB : AF
8a : 10a = 10a : AF
AF = 25/2a
AB = AF + FH + BH = 12
AB = 25/2a + 6a + 6a = 12
a = 24/49
side of square = 10a => s = 10*24/49 = 240/49
Very smart solution! 🤣
I took a quick look on the arbitrary solution for the isoceles triangle.
Let the base of triangle be *c* and the congruent legs be *a*.
Then, height h will be h = √(a²-c²/4). After a little math mumbo-jumbo one will get:
x(a,c) = [a·c·√(4a²-c²)] / [2a² + c·√(4a²-c²)], a rather complicated term.
In this case, we have c = 12, a = 10, therefore x = 10·12·√(4·10²-12²) / [2·10² + 12·√(4·10²-12²)], which gives x = 240/49 ≈ 4.89796. x² = 23.99...
*Challenge:* Where are the vertices E, D of the square lying on the lefty a-side?
How can you assume point P and point F align? There is no guarantee they are the same point.
I thought because the triangles ABC and BFG are similar triangles, you could use the angle-angle theorem
angle at point B = alpha
angle at point C = angle at point G = beta, because both are isosceles triangles
AC/AB = FG/BF
10/12 = FG/6 --> FG = 5
area square = FG² = 25
Also 12a = 6, so a = 0.5. side square = 10a = 10*0.5 = 5
What have I done wrong here ?
It seems you have assumed that F is the middle point of segment AB, which it's not.
شكرا لكم
CosACB=7/25
sinACB=24/25
sinACB=x/10-x
.....
x=240/49
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Many thanks ❤️
At a first glimpse I would say that the Area of the Pink Square is ~ 5^2 su or ~ 25 su
But if the Point F is the Middle Point between A and B, my answer is 4,8^2 su ~ 23,04 su.
I'll be back later to prove it.
You are looking great, mate. The difference is huge. In the beginning you were like "my stomach hurts, my life is terrible. I don't have any energy." Now, you are full of energy. You want to challenge yourself (exercise daily). You enjoy life. Go to Halloween parties. Don't complaint about anything. You just are bringing the masculine energy from inside and you are being a man. Keep going. Congrats.
23.999 or 24
n=4.89796, the length of the square
let the side of the square = n
the area of triangle CDG + AEF = (10-n)(n)/2 Since both have the same
base 'n' and 10-n is the difference of the side n of the square
the area of CDG + AEF = (10 n- n^2)/2
one more triangle to go: BFG
Since BFG is similar to ABC, and CF is 8 (3-4-5 triangle), then
the two legs of BFG in terms of 'n' are 8/10 (n) and 12/10 (n)
Hence, the BFG area in terms of 'n' = [ (0.8n)(1.2n)]/2 = 0.96n^2/2
Area of ABC = 12 * 9 /2 = 48, then this equal
the area of CDB + AEF, + area of the square + are of BFG = 0.96n/2
Hence 48 = (10n-n^2)/2 + n^2 + 0.96n^2/2
96 = 10 n- n^2 + 2n^2 + 0.96n^2 (multiply both sides by 2)
0 = 1.96n^2 + 10n - 96
n= 4.89796 (Quadratic formula calculator
so the length of the side of the square = n
I had first put the side of the square, but you want the area
area =4.89796 x 4.89796 = 23.9999
Great!
Many thanks ❤️
@@skwest Thanks for this comment, but I meant that
the C and "X' (the midpoint of AB ) is 8 before I looked at the Premath labeling of the diagram: ABCDEFG after working the problem on a sheet of paper. So, when I did look, I somehow used "F" for "X." Thanks again, as you are correct as
though DE lies on AF , DG does not lie on CB.
I, apparently, am the most pig-headed algebra guy, ever. I found this problem to be intensely annoying. Approached it 3× from the beginning, because I just couldn't do the algebra without being confused.
In the end though, several intersecting equations-of-lines solved this.
First, it is critical to recognize that the height of the ABC triangle is 8 units. Because … ½ of 12 base units with a hypotenuse of [10] units leaves [8] units as the rise. Pythagoras.
This in turn defines
f(𝒙) = ⁸⁄₆𝒙 ⊕ 0
f(𝒙) = ⁴⁄₃𝒙 ⊕ 0
[⊕0] because it has a (0, 0) origin. Right?
Then the diagonal “side of the square” (lower one, left) has an inverted line equation:
f'(𝒙) = -¾𝒙 + B
So what is the B? Ah … well introduce a new term [𝒂], the bit along the base, from the left which intersects with the corner of where the square meets the rising left triangle side. Because of the line function (the f(𝒙) one), we know that the height is 4𝒂/3 so
-3𝒂/4 + B = 4𝒂/3 … shift around
B = 4𝒂/3 ⊕ 3𝒂/4 … find a common denominator and combine
B = 25𝒂/12
So, that can be put into the inverted line equation
f'(𝒙) = -¾𝒙 + 25𝒂/12
We can use this immediately to find the point on the base, where it intersects.
0 = -¾𝒙 + 25𝒂/12 … rearrange to solve for 𝒙
𝒙 = 25𝒂/9
________________________________________
In a similar line of reasoning, the diagonal of the square that rises from the base to the right side is
h(𝒙) = 4𝒙/3 + B = 0 … at 𝒙 = 25𝒂/9 so
0 = ⁴⁄₃ × 25𝒂/9 + B
B = -100𝒂/27 … so
h(𝒙) = 4𝒙/3 - 100𝒂/27
𝒔² = 400𝒂²/81 … thru lots of little algebra
With that we can find where it intersects at g(𝒙), the equation for the right side of the large containing triangle.
g(𝒙) = -4𝒙/3 + 16 (because must be 0 at [𝒙 = 12])
Working through a lot of algebra setting [h(𝒙) = g(𝒙)], Icame up with
𝒙 = 6 + 75𝒂/54
Since this lower right triangle is SIMILAR to the △ABC overall one, which is isosceles, we know that
𝒔² = (6 - 75𝒂/54)² + (⁴⁄₃)²(6 - 75𝒂/54)² … reduces to
𝒔² = ²⁵⁄₉(6 - 75𝒂/54)² … thus also
400𝒂²/81 = ²⁵⁄₉(6 - 75𝒂/54)² … which expands to
441𝒂² - 48600𝒂 + 104976 = 0 … being quadratic has roots
𝒂 = [ 108 or 2.2041 ]
Clearly it cannot be the larger one, so 2.2041 ought to work.
𝒔² = 400 × 2.2041² ÷ 81
𝒔² = 23.9900
Or about 24. Yay! HARD work.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Essendo un triangolo isoscele conosco gli angoli e,con il teorema dei seni calcolo a=300/49,b=288/49,a+b=12..per cui l=300/49sin(arccos6/10)=300/49*4/5=240/49
Thanks❤️
@@PreMathBuon Natale
@@giuseppemalaguti435
Merry Christmas❤
I used another method and the area is exactly 24 cm squared, it's not circa 24 but it's exactly 24
Not so. The value is 23.99000416493127863390254060808...
@@dirklutz2818 23.990004164931 is what I got too. But my calculator isn't all that precise. LOL
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What is the meaning of your channel name sir??
Great name, isn't it. Nothing beyond Algebra 1 and Geometry on the Pre-Math channel.
I used Sin α = 0.8 to solve the side of the square EF = x.
CF / AC = EF / AF
8 / 10 = X / 6
X = 4.8
AREA = 4.8^2 = 23.04
What is wrong with my approach?
I'm curious as well, because I got same answer (only difference was getting cos  - cosines Law).
CF is not the height. F is not the middle point. AF = 6,122
@@mauriciosahady Thanks. Now I can see my mistake: it's similar, not same triangles.
𝙼𝚎𝚛𝚛𝚢 𝙲𝚑𝚛𝚒𝚜𝚝𝚖𝚊𝚜
Thank you very much!
𝙼𝚎𝚛𝚛𝚢 𝙲𝚑𝚛𝚒𝚜𝚝𝚖𝚊𝚜❤️
This topic is for advanced students so please avoid over explaining how to add the fractions. Just give the answer straight away. That will make the content more relevant
Say allaho akbar
Thanks❤️