Point in a Right Triangle

Geometry Level 4

Triangle A B C ABC is a right triangle with A B C = 9 0 \angle ABC = 90^\circ . P P is a point within triangle A B C ABC such that A P B = B P C = C P A = 12 0 \angle APB = \angle BPC = \angle CPA = 120^\circ . If P A = 15 PA = 15 and P B = 6 PB = 6 , what is the value of P C PC ?


The answer is 18.

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17 solutions

Vikram Waradpande
May 20, 2014

By the Law of Cosines, A B 2 = 6 2 + 1 5 2 2 6 15 ( 1 / 2 ) = 351 AB^2 = 6^2+15^2 - 2 * 6 * 15 * (-1/2) = 351 B C 2 = P C 2 + 6 2 2 P C 6 ( 1 / 2 ) BC^2 = PC^2 + 6^2 - 2*PC*6(-1/2) A C 2 = P C 2 + 1 5 2 2 P C 15 ( 1 / 2 ) AC^2 = PC^2 + 15^2 - 2*PC*15(-1/2) By the Pythagorean Theorem, A B 2 + B C 2 = A C 2 AB^2+BC^2=AC^2 Substitution yields 351 + P C 2 + 36 + 6 P C = P C 2 + 225 + 15 P C 351+PC^2+36+6PC=PC^2+225+15PC 162 = 9 P C P C = 18 \implies 162=9PC \implies \boxed{PC=18}

P P is known as the Fermat point of the triangle, and minimizes the sum of distances from any point to the 3 vertices.

Calvin Lin Staff - 7 years ago
Allen Jerry Aries
May 20, 2014

Solve for AB ^2 first using cosine law:

AB ^(2)= AP ^(2)+ BP ^(2)-2( AP )( BP )\cos 120^\circ AB ^(2)=15^(2)+6^(2)-2(15)(6)\cos 120^\circ AB ^(2)=351

BC ^2:

BC ^(2)= BP ^(2)+ CP ^(2)-2( BP )( CP )\cos 120^\circ BC ^(2)=6^(2)+ CP ^(2)-2(6)( CP )\cos 120^\circ BC ^(2)=36+ CP ^(2)-6( CP )

Similarly for AC ^2:

AC ^2=225+ CP ^(2)-15( CP )

By Pythagorean Theorem, AC ^2= AB ^2+ BC ^(2)

and yields PC =18

Sayantan Guha
May 20, 2014

Construct the triangle from the given data. In triangle APB, \angle APB = 120 ^ \circ So cos(120 ^ \circ) = [15^2 + 6^2 - AB^2 ] / [ 2.15.6 ] (Applying cosine rule) i.e. AB^2 = 351 Similarly applying cosine rule in triangle BPC we have BC^2 = PC^2 + 6.PC + 36 Again applying cosine rule in triangle APC we have AC^2 = PC^2 + 15.PC + 225 According to Pythagora's theorem, in triangle ABC, AB^2 + BC^2 = AC^2 Putting the obtained values- 351 + PC^2 + 6.PC + 36 = PC^2 + 15.PC + 225 Solving we get PC =18

Iskandar Setiadi
May 20, 2014

Using Law of cosines:

A B 2 = A P 2 + B P 2 2. A P . B P . c o s 12 0 AB^2 = AP^2 + BP^2 - 2.AP.BP.cos120^\circ B C 2 = B P 2 + C P 2 2. B P . C P . c o s 12 0 BC^2 = BP^2 + CP^2 - 2.BP.CP.cos120^\circ A C 2 = A P 2 + C P 2 2. A P . C P . c o s 12 0 AC^2 = AP^2 + CP^2 - 2.AP.CP.cos120^\circ

Hence,

A B 2 + B C 2 = A C 2 AB^2 + BC^2 = AC^2 A P 2 + 2 B P 2 + C P 2 + A P . B P + B P . C P = A P 2 + C P 2 + A P . C P AP^2 + 2BP^2 + CP^2 + AP.BP + BP.CP = AP^2 + CP^2 + AP.CP 2 B P 2 + A P . B P + B P . C P = A P . C P 2BP^2 + AP.BP + BP.CP = AP.CP 2. 6 2 + 15.6 + 6. C P = 15. C P 2.6^2 + 15.6 + 6.CP = 15.CP

72 + 90 = 9. C P 72 + 90 = 9.CP

C P = 162 9 = 18 CP = \frac{162}{9} = 18

Marcos Kawakami
May 20, 2014

Applying law of cosines on triangles A P B , P B C APB,PBC and A P C APC , we get:

A B 2 = P A 2 + P B 2 2 P A P B cos ( 12 0 o ) A B 2 = 351 AB^2 = PA^2 + PB^2 - 2\cdot PA\cdot PB\cdot \cos(120^{o}) \Leftrightarrow AB^2 = 351

B C 2 = P B 2 + P C 2 2 P B P C cos ( 12 0 o ) BC^2 = PB^2 + PC^2 - 2\cdot PB\cdot PC\cdot \cos(120^{o}) \Leftrightarrow B C 2 = P C 2 + 6 P C + 36 BC^2 = PC^2 + 6PC + 36 (i)

A C 2 = P A 2 + P C 2 2 P A P C cos ( 12 0 o ) AC^2 = PA^2 + PC^2 - 2\cdot PA\cdot PC\cdot \cos(120^{o}) \Leftrightarrow A C 2 = P C 2 + 15 P C + 225 AC^2 = PC^2 + 15PC + 225 (ii)

From (ii) - (i), we have

A C 2 B C 2 = 9 P C + 189 AC^2 - BC^2 = 9PC + 189

Since A B C ABC is a right triangle, we use the Pythagorean theorem to get A B 2 + B C 2 = A C 2 A C 2 B C 2 = A B 2 AB^2 + BC^2 = AC^2 \Leftrightarrow AC^2 - BC^2 = AB^2

Hence, A C 2 B C 2 = A B 2 = 351 = 9 P C + 189 P C = 18 AC^2 - BC^2 = AB^2 = 351 = 9PC + 189 \Leftrightarrow \boxed{PC = 18}

John Norman Beren
May 20, 2014

First, we get the values of sides AB, BC and AC using Cosine Law and express them in terms of PC. Since a right triangle is given, we can relate the 3 sides using Pythagorean Theorem. But, instead of determining the sides themselves, it is more convenient to look for their squares since Pythagorean Theorem states, AB^2 + BC^2 = AC^2.

Using Cosine Law for triangle APB to determine AB^2, AB^2 = PA^2 + PB^2 - 2 PA PB cos(<APB) AB^2 = 15^2 + 6^2 - 2 15 6 cos(120) AB^2 = 15^2 + 6^2 + 15*6

for Cosine Law for triangle APC to determine AC^2, AC^2 = PA^2 + PC^2 - 2 PA PC cos(<CPA) AC^2 = 15^2 + PC^2 - 2 15 PC cos(120) AC^2 = 15^2 + PC^2 + 15*PC

Cosine Law for triangle BPC to determine BC^2, BC^2 = PB^2 + PC^2 - 2 PB PC cos(<BPC) BC^2 = 6^2 + PC^2 - 2 6 PC cos(120) BC^2 = 6^2 + PC^2 + 6*PC

Using Pythagorean Theorem, AB^2 + BC^2 = AC^2 (15^2 + 6^2 + 15 6) + (6^2 + PC^2 + 6 PC) = 15^2 + PC^2 + 15*PC Solving for PC,

PC = 18 :)

Using cosine rule A B 2 AB^2 = P A 2 PA^2 + P B 2 PB^2 -2 PA PB cos 120

But Cos 120 =-0.5

So

A B 2 AB^2 = P A 2 PA^2 + P B 2 PB^2 +PA PB

Similarly

B C 2 BC^2 = P B 2 PB^2 + P C 2 PC^2 -2 PB PC cos 120

B C 2 BC^2 = P B 2 PB^2 + P C 2 PC^2 +PB PC

A C 2 AC^2 = P C 2 PC^2 + P A 2 PA^2 -2 PC PA cos 120

A C 2 AC^2 = P C 2 PC^2 + P A 2 PA^2 +PC PA

As ABC is a right angle triangle by Pythagoras theorem

P C 2 PC^2 + P A 2 PA^2 + PC PA = P B 2 PB^2 + P C 2 PC^2 +PB PC+ P A 2 PA^2 + P B 2 PB^2 +PA PB

Substitute the values for PA and PB and then solving for PC we get PC=18

Mayank Kaushik
Jul 29, 2013

By cosine rule

A B 2 = 1 5 2 + 6 2 2 15 6 cos 120 = 351 AB^{2} = 15^{2} + 6^{2} - 2*15*6*\cos120 = 351

B C 2 = 6 2 + P C 2 2 6 P C cos 120 = 36 + P C 2 + 6 P C BC^{2} = 6^{2} + PC^{2} - 2*6*PC*\cos120 = 36 + PC^{2} + 6PC

A C 2 = 1 5 2 + P C 2 2 15 P C cos 120 = 225 + P C 2 + 15 P C AC^{2} = 15^{2} + PC^{2} - 2*15*PC*\cos120 = 225 + PC^{2} + 15PC

By Pythagoreas Theorem

A B 2 + B C 2 A C 2 = 0 AB^{2} + BC^{2} - AC^{2} = 0

351 + 36 + P C 2 + 6 P C 225 P C 2 15 P C = 0 351 + 36 +PC^{2} +6PC - 225 - PC^{2} - 15PC = 0

162 9 P C = 0 162 - 9PC = 0

P C = 18 PC = 18

Moderator note:

Note: P P is also known as the Fermat point of the triangle. Do you know any other properties of it?

The Fermat point of the triangle is the point whose distance from the three vertices is minimized. Like in the problem, all of the angles are necessarily 120 degrees. It can be constructed by constructing equilateral triangles on each face using the vertices and connecting the point of the equilateral triangle not on the original triangle to the vertex opposite the side which contains it.

Ji Young Shin - 7 years, 10 months ago
Saima Ehtisham
May 20, 2014

By the cosines's law

AB^2 = 6^2+15^2 - 2 * 6 * 15 * (-1/2) = 351

AC^2 = PC^2 + 15^2 - 2 PC 15(-1/2) BC^2 = PC^2 + 6^2 - 2 PC 6(-1/2)

AC^2-BC^2 = AB^2 = 351, so

351 = 189 + 9 PC 9 PC= 162 PC = 18

Nocturne Moi
May 20, 2014

AB^2 = 6^2+15^2 - 2 * 6 * 15 * (-1/2) = 351

AC^2 = PC^2 + 15^2 - 2 PC 15(-1/2) BC^2 = PC^2 + 6^2 - 2 PC 6(-1/2)

AC^2-BC^2 = AB^2 = 351, so

351 = 189 + 9PC 9PC = 162 PC=18

Let PC = x. By Cosine Rule applied to the component triangles PAB, PBC, PAC respectively, we obtain AB^2 = 351, BC^2 = x^2 + 6x + 36, AC^2 = x^2 + 15x + 225. By Pythagoras, AC^2 = AB^2 + BC^2 . Hence x^2 + 15x + 225 = x^2 + 6x + 387, 9x = 162, x = 18.

Raphael Nasif
May 20, 2014
  • Length BA = c c ,
  • Length AC = b b ,
  • Length CB = a a ,
  • Length PC = x x

Using the cosine rule :

c 2 = 1 5 2 + 6 2 2 ( 15 ) ( 6 ) cos ( 120 ) c^2 = 15^2 + 6^2 - 2(15)(6)\cos(120)

c 2 = 351 ∴ c^2 = 351

Pythagoras' Thereom states that a 2 + b 2 = c 2 a^2 + b^2 = c^2

351 + a 2 = b 2 351 + a^2 = b^2

We can now use the cosine rule to express a 2 a^2 and b 2 b^2 in terms of x x :

b 2 = 1 5 2 + x 2 2 ( 15 ) ( x ) cos ( 120 ) b^2 = 15^2 + x^2 - 2(15)(x)\cos(120)

b 2 = 225 + x 2 + 15 x b^2 = 225 + x^2 + 15x

a 2 = 6 2 + x 2 2 ( 6 ) ( x ) cos ( 120 ) a^2 = 6^2 + x^2 - 2(6)(x)\cos(120)

a 2 = 36 + x 2 + 6 x a^2 = 36 + x^2 + 6x

Returning to Pythagoras' Thereom:

351 + a 2 = b 2 351 + a^2 = b^2

351 + 36 + x 2 + 6 x = 225 + x 2 + 15 x 351 + 36 + x^2 + 6x = 225 + x^2 +15x

9 x = 162 9x = 162

x = 18 ∴ x = 18

Vishwash Kumar
Oct 5, 2016

It iis shocking that all the 16 solvers have given the same solution although on Brilliant we are addicted to see variety in solutions ....

Ganesh Jadhav
May 20, 2014

18

Dajon Thomas
Aug 3, 2013

The key to solving this problem is using law of cosines for the three sub-triangles within the right triangle. We also know that B^2= A^2+C^2 (Pythagorean's Theorem). Thus, for Triangle APB the equation is C^2=15^2+6^2-2(15)(6)(cos(120º))=351. We should set x as PC. Then, B^2=x^2+15^2-2(15)(x)(cos(120º), and A^2=x^2+6^2-2(x)(6)(cos(120º)). To erase the x^2 term, we do B^2-A^2=C^2. This leaves 9x+189=351 and with simple arithmetic we get x (PC) is equal to 18.

Albert Ho
Aug 1, 2013

Law of Cosines Bash: A B 2 = 1 5 2 + 6 2 2 ( 15 ) ( 6 ) c o s ( 120 ) = 351 AB^{2} = 15^{2} + 6^{2} - 2(15)(6)cos(120) = 351 B C 2 = x 2 + 6 2 2 ( x ) ( 6 ) c o s ( 120 ) = x 2 + 6 x + 36 BC^{2} = x^{2} + 6^{2} - 2(x)(6)cos(120) = x^{2} + 6x + 36 A C 2 = x 2 + 1 5 2 2 ( x ) ( 15 ) c o s ( 120 ) = x 2 + 15 x + 225 AC^{2} = x^{2} + 15^{2} - 2(x)(15)cos(120) = x^{2} + 15x + 225 By Pythagorean, A B 2 + B C 2 = A C 2 AB^{2} + BC^{2} = AC^{2} , so 351 + x 2 + 6 x + 36 = x 2 + 15 x + 225 351+x^{2} + 6x + 36=x^{2} + 15x + 225 . Simplifying, we get 162 = 9 x 162=9x , so x x , which is the length of P C PC , equals 18 .

Atif Zeya
Jul 30, 2013

consider triangle APB, by cosine rule, AB^2 = AP^2 + BP^2 - 2.AP.BP.cos120 = 225+36+90. Therefore AB^2=351....(i) In triangle APC, AC^2=AP^2+PC^2-2.AP.PC.cos120....(ii), also from triangle BPC, BC^2=BP^2+PC^2-2BP.PC.cos120....(iii), and since AC^2=AB^2+BC^2(Pythagoras theorem), so equation (i) + (iii)=(ii), so, 351+36+PC^2+6PC=225+PC^2+15PC, or, 9PC=162,or PC=18

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