Points and Squares

Geometry Level 4

Let $ABC$ be a right-angled triangle at $B$ . Take points $P_1, P_2, P_3, \dots , P_7$ on $\overline {AC}$ such that $AP_1 = P_1P_2 = \dots = P_6P_7 = P_7C$ . If $BP_1^2 + BP_2^2 + \dots + BP_7^2 = 70$ , find the value of $\overline{AC}$ .

The answer is 5.656.

2 solutions

John Samuel
Mar 25, 2014

Let ABC be the right triangle, right angled at B.

Given that P1, P2, ..., P7 are points on AC and AP1 = P1P2 = ..... = P7C = P (Lets say)

Drop perpendiculars from P1, P2,...,P7 to BC, we can see that the perpendiculars divide the side AB in to 8 equal parts with length say x, such that 8x = BC

Similarly draw perpendiculars from P1, P2,...,P7 to AB, we can see that the perpendiculars divide the side AB in to 8 equal parts with length say y, such that 8y = AB

Now we can write the following
AC = 8P
BC = 8x
AB = 8y
AC^2 = AB^2 + BC^2
64P^2 = 64x^2 + 64y^2
P^2 = x^2 + y^2
BP1^2 = x^2 + 49y^2
BP2^2 = 4x^2 + 36y^2
BP3^2 = 9x^2 + 25y^2
BP4^2 = 16x^2 + 16y^2
BP5^2 = 25x^2 + 9y^2
BP6^2 = 36x^2 + 4y^2
BP7^2 = 49 x^2 + y^2

Adding up: BP1^2 + BP2^2 +.....+BP7^2 = 140(x^2 + y^2) = 70 ==> (x^2 + y^2) = 0.5

x^2 + y^2 = P^2 = 70/140=> P = sqrt(0.5)
AC = 8P = 8sqrt(0.5) = 5.65685424949238

A slightly more direct explanation is that since $P_4$ is the midpoint of $AC$ , we can repeatedly apply Stewart's Theorem, which gives us equations to simplify the sum of $BP_i ^2$ into just $BP_4 ^2$ and $AC^2$ .

Then since we have a right triangle, use $BP_4 = AP_4 = CP_4$ to solve.

Calvin Lin Staff - 7 years, 2 months ago

p4 is a midpoint only if the triangle is isoceles, (special case)

John Samuel - 7 years, 2 months ago

You are given that $AP_1 = P_1 P_2 = P_2P_3 = P_3P_4 = P_4P_5 = P_5P_6 = P_6P_7 = P_7C$ , so each of these segments has length $\frac{1}{8} AC$ , which is why $P_4$ is the midpoint. It does not make any assumption about special cases.

Calvin Lin Staff - 7 years, 2 months ago

I did that too :)

A Former Brilliant Member - 7 years, 2 months ago

this is wrong, actually AC = 4*sqrt(2) = 5.65685424949238

Chandler West - 7 years, 2 months ago

Oops... corrected my solution, went wrong when adding up

John Samuel - 7 years, 2 months ago

Jonnalagadda Mahesh
Mar 26, 2014

Consider ABC is an isosceles right angled triangle (if it holds good for right angled triangle it should hold good for an isosceles right angled triangle).

From the given condition we can conclude ${ AP }_{ 1 }={ P }_{ 1 }{ P }_{ 2 }=....={ P }_{ 6 }{ P }_{ 7 }={ P }_{ 7 }C=\frac { AC }{ 8 }$ The following triangles forms a right angled triangle at P4 ABP4 and CBP4 $\\Consider\quad triangle\quad B{ P }_{ 4 }{ P }_{ 3 }and\quad B{ P }_{ 4 }{ P }_{ 5 }\quad and\quad apply\quad pythagarous\quad theorem\\ { BP }_{ 3 }^{ 2 }={ BP }_{ 5 }^{ 2 }={ BP }_{ 4 }^{ 2 }+{ (\frac { AC }{ 8 } })^{ 2 }\\ Simillary\quad consider\quad B{ P }_{ 4 }{ P }_{ 2 },B{ P }_{ 4 }{ P }_{ 1 },B{ P }_{ 4 }{ P }_{ 6 },B{ P }_{ 4 }{ P }_{ 7 },B{ P }_{ 4 }C,B{ P }_{ 4 }A\quad and\quad we\quad get\quad the\quad following\quad results\\ \\ { BP }_{ 2 }^{ 2 }={ BP }_{ 6 }^{ 2 }={ BP }_{ 4 }^{ 2 }+{ (\frac { 2.AC }{ 8 } })^{ 2 }\\ { BP }_{ 1 }^{ 2 }={ BP }_{ 7 }^{ 2 }={ BP }_{ 4 }^{ 2 }+{ (\frac { 3.AC }{ 8 } })^{ 2 }\\ { BA }^{ 2 }={ BC }^{ 2 }={ BP }_{ 4 }^{ 2 }+{ (\frac { 4.AC }{ 8 } })^{ 2 }\\ Given\quad { BP }_{ 1 }^{ 2 }+{ BP }_{ 2 }^{ 2 }+...+{ BP }_{ 6 }^{ 2 }+{ BP }_{ 7 }^{ 2 }=70\\ Substitute\quad the\quad values\quad and\quad solve\quad u\quad will\quad get\\ \frac { 28.{ AC }^{ 2 } }{ 64 } +7.{ BP }_{ 4 }^{ 2 }=70\quad \quad \quad (Equ\quad 1)\\ { BP }_{ 4 }^{ 2 }={ BA }^{ 2 }-{ (\frac { 4.AC }{ 8 } })^{ 2 }\\ But\quad { AC }^{ 2 }={ BC }^{ 2 }+{ BA }^{ 2 }\quad \quad (BC=BA)\\ { BA }^{ 2 }=\frac { { AC }^{ 2 } }{ 2 } \\ Finally\quad { BP }_{ 4 }^{ 2 }=\frac { { AC }^{ 2 } }{ 2 } -{ (\frac { 4.AC }{ 8 } })^{ 2 }=\frac { { AC }^{ 2 } }{ 4 } \\ on\quad substituting\quad in\quad equ\quad 1\\ we\quad get\quad AC=4.\sqrt { 2 } =5.656\\ \\$

You should not use special case in your proof.

Hoo Zhi Yee - 7 years, 2 months ago

yes answer is 4*srt2 in reply to calvin lin even if p4 is mid point its not necessary that bp4p5,bp4p6 and so on will be right angled triangle as abc is not an isosceles triangle

RAHUL KUMAR CHAURASIYA - 7 years, 2 months ago

Points and Squares

The answer is 5.656.

2 solutions

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