Blue and Red are in Half!

Geometry Level 3

The rectangle above is divided into 10 squares.

What is ${\color{#3D99F6} \text{(blue angle)}} - {\color{#D61F06} \text{(red angle)}}$ in degrees?

The answer is 45.

9 solutions

Dan Ley
Apr 10, 2017

$\angle BAX=90-{\color{#3D99F6} \text{(blue angle)}}\implies \angle BAC=90$

Since lengths $\overline{AB}=\overline{AC}$ , $\triangle ABC$ is right-angled and isosceles, so its other two angles are both $45$

So, ${\color{#3D99F6} \text{(blue angle)}}-{\color{#D61F06} \text{(red angle)}}=\angle ABC = 45$

Moderator note:

While it's a common technique to add lines to a geometric figure to aid in the solution, the grid for the original problem adds an extra barrier to realizing this. It's easy to get "locked in" and not realize the "graph paper" the problem uses can extend in any direction.

Great job. Thinking a little more, this construction shows the key fact is that the red triangle has legs of 3+2=5 and 3-2=1 (based on the blue triangle's legs of 3 and 2).

You could make similar problems/constructions to this one with sides x and y for the blue triangle and sides x+y and x-y for the red triangle.

Matthew Feig - 4 years, 2 months ago

Right! I see where you're coming from, it's cool that we have the general case now. Maybe I should have included the red triangle with hypotenuse $\overline{BC}$ in the image.

Dan Ley - 4 years, 2 months ago

Elegant solution without trigonometry.

Can you explain what motivated you to try this construction?

Agnishom Chattopadhyay - 4 years, 2 months ago

Trig was instantly obvious, but much less exciting. I needed to find a way to represent the blue angle minus the red angle on the diagram, and extending B to C seemed to work- then it was just a matter of spotting things. It's been used in the above method too, but I thought the diagram in context would help too.

Dan Ley - 4 years, 2 months ago

Wow. Your explanation doesn't exactly outline the rationale behind your statement, but your solution is awesome.

Barak-Har Elkin - 4 years, 1 month ago

Thanks, which part of the explanation is unclear?

Dan Ley - 4 years, 1 month ago

I think Barak-Har Elkin would find your explanation of the solution spot on if, as you discussed above with Matthew Feig, your diagram included the squares behind the triangle with hypotenuse BC, and you included the red angle there. Then the brilliance of the isosceles triangle ABC illuminating the subtraction of blue and red becomes visually evident. One final addition? a right angle indicator at the vertex of BAC. Then you hardly need words at all! :D

Andrew Deller - 4 years, 1 month ago

I thought inverse trig straight away. This is a beautiful solution!

Surya Teja Chavali - 4 years, 1 month ago

Brian Charlesworth
Apr 3, 2017

Relevant wiki: Basic Trigonometric Functions

Let $\angle B$ be the blue angle and $\angle R$ be the red angle. Then $\angle B = \tan^{-1}(\frac{3}{2})$ and $\angle R = \tan^{-1}(\frac{1}{5})$ , and so

$\angle B - \angle R = \tan^{-1}(\frac{3}{2}) - \tan^{-1}(\frac{1}{5}) = \tan^{-1}\left(\dfrac{\frac{3}{2} - \frac{1}{5}}{1 + \frac{3}{2} \times \frac{1}{5}}\right) = \tan^{-1}\left(\dfrac{\frac{13}{10}}{\frac{13}{10}}\right) = \tan^{-1}(1) = \boxed{45^{\circ}}$ .

Smooth application of trigonometry. For someone without the knowledge of trig, this problem seems quiet hard (although not impossible)

Agnishom Chattopadhyay - 4 years, 2 months ago

Plz explain how tan inverse method use in

Tamil Tamil - 4 years, 1 month ago

Starting with the familiar formula

$\tan(a - b) = \dfrac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)}$ ,

we then have that $a - b = \arctan\left(\dfrac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)}\right)$ .

Then letting $a = \arctan(A), b = \arctan(B)$ we end up with

$\arctan(A) - \arctan(B) = \arctan\left(\dfrac{A - B}{1 + AB}\right)$ .

There are some restrictions on when the formula can be used but this is the general idea.

Brian Charlesworth - 4 years, 1 month ago

The same way man! (+1)

Noel Lo - 3 years, 11 months ago

That isn't correct tangent is rise over run in that solution it is showing run over rise with the 3/2 and ride over run with the 1/5

Ben Wood - 4 years, 2 months ago

Tangent is better thought of as Opposite over Adjacent than rise over run. You will be able to see the correct sides to use more easily if you use that definition rather than the definition for slope. The rise over run only applies if a side touching the angle is on the x-axis.

Alex Li - 4 years, 2 months ago

Ahmad Saad
Apr 4, 2017

How do you know that $\angle{BAD} = 90^\circ$ ?

Christopher Boo - 4 years, 2 months ago

<DAE = <ABF , <BAF + <ABF = 90 deg.

<BAD = <BAF + <DAE = <BAF + ABF = 90 deg.

Ahmad Saad - 4 years, 2 months ago

Marcus Neal
Apr 14, 2017

The blue angle can be represented as the argument of the complex number 2+3i, and likewise the red angle as the argument of 5+i. Their difference would be the argument of (2+3i)/(5+i) = (13+13i)/26, which is 45 degrees.

Ah, solving this problem via complex numbers. An underrated approach! This brings me back to my high school years when I proved many Machin-like formula using this very method.

Good read!

Pi Han Goh - 4 years, 1 month ago

Good solution hahahaah cool

Maju Sumanto Tampubolon - 4 years, 1 month ago

A Former Brilliant Member
Apr 3, 2017

Let the $\color{#3D99F6}blue~angle$ be $\color{#3D99F6}\theta$ , and the $\color{#D61F06}red~angle$ be $\color{#D61F06}\phi$ . Assuming the the side length of each square is $1$ unit, we have

$tan\theta=\frac{3}{2}$ $\implies$ $\color{#3D99F6}\theta=56.30993247$

$tan\phi=\frac{1}{5}$ $\implies$ $\color{#D61F06}\phi=11.30993247$

Finally,

$\color{#3D99F6}\theta - \color{#D61F06}\phi$ $= \color{#3D99F6}56.30993247$ $-$ $\color{#D61F06}11.30993247$ $=$ $\boxed{\color{#20A900}\large45}$

This method only shows that $\theta - \phi$ is approximately equal to 45. The first 8 decimal places of $\theta$ and $\phi$ are the same, but it is possible that they may have different values further on.

This is quite similar to first problem of this week's intermediate set, where the two numbers have the same three digits after the decimal point but are not equal to each other.

Pranshu Gaba - 4 years, 1 month ago

Roger Erisman
Apr 3, 2017

tan(blue) = 3/2

tan(red) = 1/5

blue = arctan(3/2) = 56.3 degrees

red = arctan(0.2) = 11.3 degrees

blue-red = 45 degrees

This method only shows that (blue - red) is approximately equal to 45. The first decimal place of blue and red are the same, but it is possible that they may have different values further on.

This is quite similar to first problem of this week's intermediate set, where the two numbers have the same three digits after the decimal point but are not equal to each other.

Pranshu Gaba - 4 years, 1 month ago

The problem only accepted integer solutions

A Former Brilliant Member - 4 years, 1 month ago

That doesn't mean that the answer has to be an integer. It is the responsibility of the solution writer to prove that the answer is exactly 45. This solution only says that "The answer is 45 because well, I can't submit a non-integer answer."

Pi Han Goh - 4 years, 1 month ago

@Pi Han Goh – The calculation shows that the answer is approximately 45, the problem only needed an approximate solution. As asked and answered, this calculation is sufficient to solve the problem.

A Former Brilliant Member - 4 years, 1 month ago

@A Former Brilliant Member – The question asked to find the exact solution, not "round your answer to the nearest integer" solution.

Pi Han Goh - 4 years, 1 month ago

David Zak
Apr 13, 2017

angle YCB = ZBC = DAE (red) (triangles CYB, ZBC, AED congruent)
angle ACX = BAF = (90 - ABF) = (90 - blue) (sum of angles in triangle BAF = 180)
AC = AB (triangles AXC, BFA congruent)
Therefore angle ACB = ABC (isosceles triangle ABC); ACX + YCB = ABF - CBF; (90 - blue + red) = (blue - red)
Thus, 2(blue - red) = 90 and (blue - red) = 45 degrees.

Oh, a nice variation of Dan Ley's approach! This certainly works too~

Pi Han Goh - 4 years, 1 month ago

Prayas Rautray
Apr 5, 2017

Tan b= 3/2 Tan r= 1/5 Tan(b-r)=(tan b-tan r)/1+(tan b)(tan r) ={(3/2)-(1/5)}/{1+(3/10)} =1 So, b-r=45° b is blue and r is red

For completeness, can you explain why $\tan(b-r) = 1$ implies $b-r = 45^\circ$ only? Why can't $b-r = 225^\circ$ ?

Pi Han Goh - 4 years, 1 month ago

Angel Krastev
Apr 13, 2017

It is 45 degrees. Lets find tg(blue-red): tg(b-r)=(tg(b)-tg(r))/(1+tg(b) tg(r);
tg(b)=3/2; tg(r)=1/5 (1,5-0.2)/(1+1.5
0.2) = 1.3/1.3 = 1. If tg(b-r)=1 then b-r=45 degrees.

Blue and Red are in Half!

The answer is 45.

9 solutions

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