Pink Valentine's Day Triangle
In the following picture, there are three squares of side length 1.
Find the area of the pink triangle.
Source: University of Lincoln 2015.
The answer is 0.8.
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12 solutions
Nice solution, I did the same.
I used the same method... Really, understanding geometry through algebra is a powerful tool.
The area of the pink triangle can be written as...
ΔABI = ( ΔABF - ( ΔBFH - ( ΔFHJ - ( ΔHJK - ............ ∞ ) ) ) )
ΔABI = ( 1 - ( 4 1 - ( 1 6 1 - ( 6 4 1 - ......∞ ) ) ) )
ΔABI = 1 - 4 1 + 1 6 1 - 6 4 1 + ......∞
This is an infinite geometric series.
Let a=1, Common ratio r = - 4 1 .
ΔABI = 1 − r a = 1 − ( − 1 / 4 ) 1 = 5 4
Area of pink triangle (ΔABI) = 0.8
Nice solution. Its cool.
Nicest solution ever
It can also be done using trigonometry
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yep... if the area of each square is 1, the squares must be 1 long.
Considering the triangle ABD, a right angled triangle with BD 1 long and AB 2 long, tan A = 0.5 therefore A = 26.57 degrees . It can be seen that BE extending to the right hand side of the L shape intersects AD at right angles.
ABC is a right angled triangle with the hypotenuse AB, which is 2 long. Since we know A = 26.57 degrees and sin 26.57 = 0.4472 we can determine BC as 0.8944, and AC as 1.7889.
The area of ABC is therefore 0.5 x 0.8944 x 1.7888 = 0.8
(sorry, no luck doing the LaTeX tags!)
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How BC can be 1 when BD is 1.
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@Muhammad Arslan Maan – When Andy is being thick and mixing up his sins and tans... now fixed.
yes, exactly...
Alternative answer not using multiple significant figures...
For ABD, we know BD = 1 and AB = 2, therefore AD = 5^0.5
For BCD, we know BD = 1, and the internal angles are the same as ABD - angle B of BCD is the same as angle A of ABD - so cos B = 2/5^0.5. Taking cos B = BC/BD, cos B = BC/1 and therefore BC = 2/5^0.5.
To calculate AC, take AB^2 = AC^2 + BC^2
2^2 = AC^2 + 4/5
16/5 = AC^2
AC = 4/5^0.5
and finally, the area of ABC = (0.5)(BC)(AC) = (0.5)(2/5^0.5)(4/5^0.5) = 4/(5^0.5)(5^0.5) = 4/5 = 0.8
Actually, I found a completely different solution avoiding trigonometry.
The area of triangle ABD is 1, and its hypotenuse (AD) is 5^0.5
If we take the hypotenuse as the base of ABD, the height (BC) is calculated as:
Area = (0.5)(BC)(AD) 1 = (0.5)(BC)(5^0.5) 2/5^0.5 = BC
AB^2 = BC^2 + AC^2, therefore: 2^2 = 2/(5^0.5)^2 + AC^2 4 = 4/5 + AC^2 16/5 = AC^2 AC = 4/5^0.5
Area of ABC is therefore (1/2)(2/5^0.5)(4/5^0.5) = 8/10 = 0.8
Alternatively, as someone else posted, the hypotenuse of ABD is 5^0.5 and the hypotenuse of ABC is 2, and they are similar triangles.
If the hypotenuses are in the ratio 2:5^0.5 then the areas are in the ratio 4:5, or 0.8.
I did the same.
The pink triangle and the bottom triangle are similar with hypotenuse 2 and 1 respectively. As such, the length of their sides are in the ratio of 2:1 and therefore, the areas in the ratio of 4:1. As the combined area of these two triangles is 1, the area of pink triangle would be 4/5.
Wow, it's very sad that this isn't the top solution!
Ha, I did that, too! Well, I compared the bases, anyway.
I did exactly the same. Quite simple!
2 1 2 2 S i n ( A r c T a n ( 2 1 ) ) C o s ( A r c T a n ( 2 1 ) ) = 5 4
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2nd approach, subtracting similar triangle from larger 1 × 2 triangle
( 2 1 1 ⋅ 2 ) − ( 1 2 + 2 2 1 ) 2 ( 2 1 1 ⋅ 2 ) = 5 4
Great solution. I didn't think of this.
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Such simple task must not to be found here!!!
I integrated (-2x+2-x/2)dx from 0 to 4/5 and I obtained the correct answer: 4/5.
Comparing pink to green:
Ratio of hypotenuses: 2: 5
Ratio of areas: 4:5
Area of green: 1
Area of pink: 4/5
All 3 triangles ABC, CDE and FDG are right triangles, and so must be similar if they also share one other angle. Angles c and b are clearly equal. Angles a and b are also equal as AB and EF are parallel. Therefore ABC and DFG are similar and so ∣ A C ∣ ∣ B C ∣ = ∣ F G ∣ ∣ F D ∣ = 0.5. The length of the hypotenuse of ABC is 2 and so by Pythagoras |BC| = 5 2 and the area of ABC is 0.8
Let B be the origin, B A lie along the positive y -axis and the positive x -axis lie to the right of B .
Then B C lies along the line y = 2 1 x and C A lies along the line y − 2 = − 2 x ⟹ y = − 2 x + 2 .
The point of intersection of these two lines is then C , the x -coordinate of which is such that
2 1 x = − 2 x + 2 ⟹ 2 5 x = 2 ⟹ x = 5 4 .
With B A being the base of Δ A B C , this triangle then has a base of length 2 and a height of 5 4 , and thus has an area of 2 1 ∗ 2 ∗ 5 4 = 5 4 = 0 . 8 .