2021 AIME I Problems/Problem 6: Difference between revisions

Revision as of 13:00, 14 March 2021

Problem

Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},$ and $PG=36\sqrt{7}$ . What is $PA$ ?

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=vaRfI0l4s_8

Solution 1

First scale down the whole cube by 12. Let point P have coordinates $(x, y, z)$ , A have coordinates $(0, 0, 0)$ , and $s$ be the side length. Then we have the equations $(s-x)^2+y^2+z^2=(5\sqrt{10})^2$ $x^2+(s-y)^2+z^2=(5\sqrt{5})^2$ $x^2+y^2+(s-z)^2=(10\sqrt{2})^2$ $(s-x)^2+(s-y)^2+(s-z)^2=(3\sqrt{7})^2$ These simplify into $s^2+x^2+y^2+z^2-2sx=250$ $s^2+x^2+y^2+z^2-2sy=125$ $s^2+x^2+y^2+z^2-2sz=200$ $3s^2-2s(x+y+z)+x^2+y^2+z^2=63$ Adding the first three equations together, we get $3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575$ . Subtracting this from the fourth equation, we get $2(x^2+y^2+z^2)=512$ , so $x^2+y^2+z^2=256$ . This means $PA=16$ . However, we scaled down everything by 12 so our answer is $16*12=\boxed{192}$ . ~JHawk0224

Solution 2 (Solution 1 with slight simplification)

Once the equations for the distance between point P and the vertices of the cube have been written. We can add the first, second, and third to receive, $2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.$ Subtracting the fourth equation gives, $2(x^2 + y^2 + z^2) = 575 - 63$ $x^2 + y^2 + z^2 = 256$ $\sqrt{x^2 + y^2 + z^2} = 16.$ Since point $A = (0,0,0), PA = 16$ , and since we scaled the answer is $16 \cdot 12 = \boxed{192}$ ~Aaryabhatta1

Solution 3

Let E be the vertex of the cube such that ABED is a square. By the British Flag Theorem, we can easily we can show that $PA^2 + PE^2 = PB^2 + PD^2$ and $PA^2 + PG^2 = PC^2 + PE^2$ Hence, adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$ . Substituting in the values we know, we get $2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2$ .

Thus, we can solve for $PA$ , which ends up being $\boxed{192}$ .

(Lokman GÖKÇE)

@@ Line 25: / Line 25: @@
 ==Solution 3==
-By Pythagorean Theorem, easly we can show that
+Let E be the vertex of the cube such that ABED is a square.
+By the British Flag Theorem, we can easily we can show that
 <cmath>PA^2 + PE^2 = PB^2 + PD^2</cmath>
+and
 <cmath>PA^2 + PG^2 = PC^2 + PE^2</cmath>
-Hence, <math>2PA^2 + PG^2 = PB^2 + PC^2 + PD^2</math>. <math>2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2 </math>.
+Hence, adding the two equations together, we get <math>2PA^2 + PG^2 = PB^2 + PC^2 + PD^2</math>. Substituting in the values we know, we get <math>2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2 </math>.
-Thus <math>PA</math> is <math>\boxed{192}</math>.
+Thus, we can solve for <math>PA</math>, which ends up being <math>\boxed{192}</math>.
 (Lokman GÖKÇE)

2021 AIME I (Problems • Answer Key • Resources)
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