Well, i don't think this is too easy

Geometry Level 4

Given in the figure is a 20 foot ladder leaning against a wall, at the bottom is a 6 by 6 by 6 cube touching the ladder at one point and pushed against the bottom of the wall. You need to solve for the height marked in the figure. If height is 'H', enter your answer as [H] {where [x] is the floor function}

I did not make this problem and is taken from here


The answer is 17.

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2 solutions

Satyen Nabar
Aug 8, 2014

A different approach without trigonometry---

Let ABC be the triangle with angle B 90 degrees. Let BDEF be the square with D on AB (top of cube on wall) and F on BC (cube bottom)

Let x be the length of line segment AD, and let y be the length of line segment CF.

1)Because of the similarity of the triangles ADE and EFC, the following holds: x:6=6:y

So xy=36 . Also x=36/y and y =36/x

2)By Pythagoras,

(x+6)^2 + (y+6)^2 = 20^2

(x^2+72+y^2) + 12x+12y=400

Using xy=36,

(x^2+2xy+y^2) +12x +12y=400

(x+y)^2 +12(x+y)-400 =0.

Solving quadratic equation (x+y) = 14.88

3) x=36/y so x+(36/x)= 14.88

x^2+36= 14.88x

x^2-14.88x+36=0

Solving quadratic equation,

x= 11.83 or 3.04

The height of wall where ladder touches is 11.83 + 6 = 17.83

Floor function = 17.

nice :) ........ this is what i was talking about in the discussion of the other solution

Abhinav Raichur - 6 years, 10 months ago

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shit, i didn't write the floor function. i wrote the height

Cisco Wicaksana - 5 years, 6 months ago
Ronak Agarwal
Jul 20, 2014

Well that seems to be easy and it it easy. Trigo Trigo N o t e t h a t B E = 20 c o s ( θ ) a n d E C = F C = 6 a n d B C = E B E C = 20 c o s ( θ ) 6 I n t r i a n g l e B C F w e h a v e C F B C = t a n θ p u t t i n g v a l u e s w e h a v e t a n θ = 6 20 c o s θ 6 A l l i s l e f t t o s o l v e f o r θ . 10 s i n θ 3 t a n θ = 3 5 s i n 2 θ = 3 ( s i n θ + c o s θ ) 5 c o s ( 2 θ + π 2 ) = 3 2 s i n ( θ + π 4 ) P u t θ + π 4 = x O u r e q u a t i o n b e c o m e s 5 c o s ( 2 x ) = 3 2 s i n ( x ) 10 s i n 2 ( x ) 3 2 s i n ( x ) 5 = 0 s i n ( x ) = ( 3 2 + 218 20 ) x = sin 1 ( 3 2 + 218 20 ) 71.87 0 o r 108.12 0 θ 26.87 0 o r 63.12 0 . h = 20 s i n θ 9.04 o r 17.84. h = 9 o r 17 , I s t i l l d o n t k n o w w h y 9 i s n o t t h e a n s w e r . Note\quad that\quad BE=20cos(\theta )\quad and\quad EC=FC=6\quad and\\ BC=EB-EC=20cos(\theta )-6\quad \\ In\quad triangle\quad BCF\quad we\quad have\\ \frac { CF }{ BC } =tan\theta \quad putting\quad values\quad we\quad have\\ tan\theta =\frac { 6 }{ 20cos\theta -6 } \quad All\quad is\quad left\quad to\quad solve\quad for\quad \theta .\\ \Rightarrow 10sin\theta -3tan\theta =3\quad \Rightarrow 5sin2\theta =3(sin\theta +cos\theta )\\ \Rightarrow -5cos(2\theta +\frac { \pi }{ 2 } )=3\sqrt { 2 } sin(\theta +\frac { \pi }{ 4 } )\quad \\ Put\quad \theta +\frac { \pi }{ 4 } =x\quad Our\quad equation\quad becomes\\ -5cos(2x)=3\sqrt { 2 } sin(x)\\ \Rightarrow 10{ sin }^{ 2 }(x)-3\sqrt { 2 } sin(x)-5=0\\ \Rightarrow sin(x)=(\frac { 3\sqrt { 2 } +\sqrt { 218 } }{ 20 } )\quad \\ \Rightarrow x=\sin ^{ -1 }{ (\frac { 3\sqrt { 2 } +\sqrt { 218 } }{ 20 } ) } \approx { 71.87 }^{ 0 }\quad or\quad { 108.12 }^{ 0 }\\ \Rightarrow \theta \approx { 26.87 }^{ 0 }\quad or\quad { 63.12 }^{ 0 }\quad .\\ \Rightarrow h=20sin\theta \approx 9.04\quad or\quad 17.84.\\ \left\lfloor h \right\rfloor =9\quad or\quad 17\quad ,\quad I\quad still\quad don't\quad know\quad why\quad 9\quad is\quad not\quad the\quad \\ answer.

9.04052... 9.04052... is the other answer, because the floor could be the wall, while the wall is the floor. In other words, flip the ladder. The figure is already provided.

Michael Mendrin - 6 years, 10 months ago

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right!!. :) ..... 71.87 and 108.12 are complementary angles

Abhinav Raichur - 6 years, 10 months ago

i get 9 wrong wtf

math man - 6 years, 9 months ago

it is easy, of course, when you use trigonometry :p ........ but i had tried this 2 years back without even the least conception of trigonometry!!........ hence, the name (try without trigonometry, you will enjoy it!)

Abhinav Raichur - 6 years, 10 months ago

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Well, of course, H H can be solved directly with this equation:

( H 6 ) 2 + 6 2 + 6 2 + ( 6 2 H 6 ) 2 = 20 \sqrt { { (H-6) }^{ 2 }+{ 6 }^{ 2 } } +\sqrt { { 6 }^{ 2 }+(\dfrac { { 6 }^{ 2 } }{ H-6 } )^{ 2 } } =20

which leads to the same results.

Michael Mendrin - 6 years, 10 months ago

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Here's your solution without trignomerty

Ronak Agarwal - 6 years, 10 months ago

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@Ronak Agarwal Ronak, the equation may look straightforward, but it's tedious to get at the roots, so doing it by trigonometry might be a more expedient route to a numerical answer.

Michael Mendrin - 6 years, 10 months ago

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@Michael Mendrin I know that I just told it easy because the solving part can be left to wolfram alpha.

Ronak Agarwal - 6 years, 10 months ago

@Michael Mendrin YES !... i was stuck there ......... its tricky! :)

Abhinav Raichur - 6 years, 10 months ago

whoa! yes ........ the same :p

Abhinav Raichur - 6 years, 10 months ago

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