Triangle A B C is a right triangle with ∠ A B C = 9 0 ∘ . P is a point within triangle A B C such that ∠ A P B = ∠ B P C = ∠ C P A = 1 2 0 ∘ . If P A = 1 5 and P B = 6 , what is the value of P C ?
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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
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
Using Law of cosines:
A B 2 = A P 2 + B P 2 − 2 . A P . B P . c o s 1 2 0 ∘ B C 2 = B P 2 + C P 2 − 2 . B P . C P . c o s 1 2 0 ∘ A C 2 = A P 2 + C P 2 − 2 . A P . C P . c o s 1 2 0 ∘
Hence,
A B 2 + B C 2 = A C 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 2 B P 2 + A P . B P + B P . C P = A P . C P 2 . 6 2 + 1 5 . 6 + 6 . C P = 1 5 . C P
7 2 + 9 0 = 9 . C P
C P = 9 1 6 2 = 1 8
Applying law of cosines on triangles A P B , P B C and A P C , we get:
A B 2 = P A 2 + P B 2 − 2 ⋅ P A ⋅ P B ⋅ cos ( 1 2 0 o ) ⇔ A B 2 = 3 5 1
B C 2 = P B 2 + P C 2 − 2 ⋅ P B ⋅ P C ⋅ cos ( 1 2 0 o ) ⇔ B C 2 = P C 2 + 6 P C + 3 6 (i)
A C 2 = P A 2 + P C 2 − 2 ⋅ P A ⋅ P C ⋅ cos ( 1 2 0 o ) ⇔ A C 2 = P C 2 + 1 5 P C + 2 2 5 (ii)
From (ii) - (i), we have
A C 2 − B C 2 = 9 P C + 1 8 9
Since A B C 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
Hence, A C 2 − B C 2 = A B 2 = 3 5 1 = 9 P C + 1 8 9 ⇔ P C = 1 8
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 = P A 2 + P B 2 -2 PA PB cos 120
But Cos 120 =-0.5
So
A B 2 = P A 2 + P B 2 +PA PB
Similarly
B C 2 = P B 2 + P C 2 -2 PB PC cos 120
B C 2 = P B 2 + P C 2 +PB PC
A C 2 = P C 2 + P A 2 -2 PC PA cos 120
A C 2 = P C 2 + P A 2 +PC PA
As ABC is a right angle triangle by Pythagoras theorem
P C 2 + P A 2 + PC PA = P B 2 + P C 2 +PB PC+ P A 2 + P B 2 +PA PB
Substitute the values for PA and PB and then solving for PC we get PC=18
By cosine rule
A B 2 = 1 5 2 + 6 2 − 2 ∗ 1 5 ∗ 6 ∗ cos 1 2 0 = 3 5 1
B C 2 = 6 2 + P C 2 − 2 ∗ 6 ∗ P C ∗ cos 1 2 0 = 3 6 + P C 2 + 6 P C
A C 2 = 1 5 2 + P C 2 − 2 ∗ 1 5 ∗ P C ∗ cos 1 2 0 = 2 2 5 + P C 2 + 1 5 P C
By Pythagoreas Theorem
A B 2 + B C 2 − A C 2 = 0
3 5 1 + 3 6 + P C 2 + 6 P C − 2 2 5 − P C 2 − 1 5 P C = 0
1 6 2 − 9 P C = 0
P C = 1 8
Note: 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.
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
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.
Using the cosine rule :
c 2 = 1 5 2 + 6 2 − 2 ( 1 5 ) ( 6 ) cos ( 1 2 0 )
∴ c 2 = 3 5 1
Pythagoras' Thereom states that a 2 + b 2 = c 2
3 5 1 + a 2 = b 2
We can now use the cosine rule to express a 2 and b 2 in terms of x :
b 2 = 1 5 2 + x 2 − 2 ( 1 5 ) ( x ) cos ( 1 2 0 )
b 2 = 2 2 5 + x 2 + 1 5 x
a 2 = 6 2 + x 2 − 2 ( 6 ) ( x ) cos ( 1 2 0 )
a 2 = 3 6 + x 2 + 6 x
Returning to Pythagoras' Thereom:
3 5 1 + a 2 = b 2
3 5 1 + 3 6 + x 2 + 6 x = 2 2 5 + x 2 + 1 5 x
9 x = 1 6 2
∴ x = 1 8
It iis shocking that all the 16 solvers have given the same solution although on Brilliant we are addicted to see variety in solutions ....
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.
Law of Cosines Bash: A B 2 = 1 5 2 + 6 2 − 2 ( 1 5 ) ( 6 ) c o s ( 1 2 0 ) = 3 5 1 B C 2 = x 2 + 6 2 − 2 ( x ) ( 6 ) c o s ( 1 2 0 ) = x 2 + 6 x + 3 6 A C 2 = x 2 + 1 5 2 − 2 ( x ) ( 1 5 ) c o s ( 1 2 0 ) = x 2 + 1 5 x + 2 2 5 By Pythagorean, A B 2 + B C 2 = A C 2 , so 3 5 1 + x 2 + 6 x + 3 6 = x 2 + 1 5 x + 2 2 5 . Simplifying, we get 1 6 2 = 9 x , so x , which is the length of P C , equals 18 .
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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By the Law of Cosines, A B 2 = 6 2 + 1 5 2 − 2 ∗ 6 ∗ 1 5 ∗ ( − 1 / 2 ) = 3 5 1 B C 2 = P C 2 + 6 2 − 2 ∗ P C ∗ 6 ( − 1 / 2 ) A C 2 = P C 2 + 1 5 2 − 2 ∗ P C ∗ 1 5 ( − 1 / 2 ) By the Pythagorean Theorem, A B 2 + B C 2 = A C 2 Substitution yields 3 5 1 + P C 2 + 3 6 + 6 P C = P C 2 + 2 2 5 + 1 5 P C ⟹ 1 6 2 = 9 P C ⟹ P C = 1 8