Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1980 USAMO Problems/Problem 4"

(→‎Solution: breaking up a giant block of text)
m (→‎Solution: typo)
 
Line 5: Line 5:
 
Let <math>ABCD</math> be the tetrahedron, and let <math>P</math> and <math>Q</math> be the points at which the insphere touches faces <math>ABC</math> and <math>ABD</math> respectively (and therefore the centroids of those faces). Looking at the plane containing <math>P</math>, <math>A</math>, and <math>Q</math>, we see that the intersection of the sphere and the plane is a circle, and that <math>AP</math> and <math>AQ</math> are both tangent to said circle. <math>[AP]</math> and <math>[AQ]</math> are tangents to the circle from the same point and thus have the same length. The same goes for <math>[BP]</math> and <math>[BQ]</math>. Thus, triangles <math>ABP</math> and <math>ABQ</math> are congruent, and <math>\mathbf{\angle PAB = \angle QAB}</math>.
 
Let <math>ABCD</math> be the tetrahedron, and let <math>P</math> and <math>Q</math> be the points at which the insphere touches faces <math>ABC</math> and <math>ABD</math> respectively (and therefore the centroids of those faces). Looking at the plane containing <math>P</math>, <math>A</math>, and <math>Q</math>, we see that the intersection of the sphere and the plane is a circle, and that <math>AP</math> and <math>AQ</math> are both tangent to said circle. <math>[AP]</math> and <math>[AQ]</math> are tangents to the circle from the same point and thus have the same length. The same goes for <math>[BP]</math> and <math>[BQ]</math>. Thus, triangles <math>ABP</math> and <math>ABQ</math> are congruent, and <math>\mathbf{\angle PAB = \angle QAB}</math>.
  
−
Let the intersection of <math>AP</math> and <math>BC</math> be <math>M</math>, and let the intersection of <math>AQ</math> and <math>BD</math> be <math>N</math>. Then <math>AM</math> and <math>AN</math> are medians of <math>ABC</math> and <math>ABD</math>, and thus <math>|AM| = \frac{3}{2}|AP| = \frac{3}{2}|AQ| = |AN|</math>. We already know from the previous congruence that <math>\angle{MAB} = \angle{PAB} = \angle{QAB} = \angle{NAB}</math>, and <math>|AB|</math> is equal to itself. Thus, <math>MAB</math> and <math>NAB</math> are also congruent to each other. Finally, <math>|BC| = 2|BM| = 2|BN| = |BD|</math> (Because <math>M</math> and <math>N</math> are midpoints of <math>BC</math> and <math>BD</math> respectively), and from the congruence of <math>MAB</math> and <math>NAB</math> we have <math>\angle{CBA} = \angle{MBA} = \angle{NBA} = \angle{DBA}</math>, and again <math>|AB|</math> is equal to itself. Thus <math>ABC</math> and <math>ABD</math> are congruent, thus <math>|mathbf{|AC| = |AD|}</math> and <math>\mathbf{|BC| = |BD|}</math>.
+
Let the intersection of <math>AP</math> and <math>BC</math> be <math>M</math>, and let the intersection of <math>AQ</math> and <math>BD</math> be <math>N</math>. Then <math>AM</math> and <math>AN</math> are medians of <math>ABC</math> and <math>ABD</math>, and thus <math>|AM| = \frac{3}{2}|AP| = \frac{3}{2}|AQ| = |AN|</math>. We already know from the previous congruence that <math>\angle{MAB} = \angle{PAB} = \angle{QAB} = \angle{NAB}</math>, and <math>|AB|</math> is equal to itself. Thus, <math>MAB</math> and <math>NAB</math> are also congruent to each other. Finally, <math>|BC| = 2|BM| = 2|BN| = |BD|</math> (Because <math>M</math> and <math>N</math> are midpoints of <math>BC</math> and <math>BD</math> respectively), and from the congruence of <math>MAB</math> and <math>NAB</math> we have <math>\angle{CBA} = \angle{MBA} = \angle{NBA} = \angle{DBA}</math>, and again <math>|AB|</math> is equal to itself. Thus <math>ABC</math> and <math>ABD</math> are congruent, thus <math>\mathbf{|AC| = |AD|}</math> and <math>\mathbf{|BC| = |BD|}</math>.
  
 
By applying the same logic to faces <math>BCD</math> and <math>ACD</math> we get <math>|AC| = |BC|</math> and <math>|AD| = |BD|</math>. Finally, applying the same logic to faces <math>ACB</math> and <math>ACD</math> we get <math>|AB| = |AD|</math> and <math>|CB| = |CD|</math>. Putting all these equalities together, we get that all edges of the tetrahedron are equal, and thus the tetrahedron is regular. <math>\blacksquare</math>
 
By applying the same logic to faces <math>BCD</math> and <math>ACD</math> we get <math>|AC| = |BC|</math> and <math>|AD| = |BD|</math>. Finally, applying the same logic to faces <math>ACB</math> and <math>ACD</math> we get <math>|AB| = |AD|</math> and <math>|CB| = |CD|</math>. Putting all these equalities together, we get that all edges of the tetrahedron are equal, and thus the tetrahedron is regular. <math>\blacksquare</math>

Latest revision as of 19:12, 20 March 2023

Problem

The insphere of a tetrahedron touches each face at its centroid. Show that the tetrahedron is regular.

Solution

Let $ABCD$ be the tetrahedron, and let $P$ and $Q$ be the points at which the insphere touches faces $ABC$ and $ABD$ respectively (and therefore the centroids of those faces). Looking at the plane containing $P$, $A$, and $Q$, we see that the intersection of the sphere and the plane is a circle, and that $AP$ and $AQ$ are both tangent to said circle. $[AP]$ and $[AQ]$ are tangents to the circle from the same point and thus have the same length. The same goes for $[BP]$ and $[BQ]$. Thus, triangles $ABP$ and $ABQ$ are congruent, and $\mathbf{\angle PAB = \angle QAB}$.

Let the intersection of $AP$ and $BC$ be $M$, and let the intersection of $AQ$ and $BD$ be $N$. Then $AM$ and $AN$ are medians of $ABC$ and $ABD$, and thus $|AM| = \frac{3}{2}|AP| = \frac{3}{2}|AQ| = |AN|$. We already know from the previous congruence that $\angle{MAB} = \angle{PAB} = \angle{QAB} = \angle{NAB}$, and $|AB|$ is equal to itself. Thus, $MAB$ and $NAB$ are also congruent to each other. Finally, $|BC| = 2|BM| = 2|BN| = |BD|$ (Because $M$ and $N$ are midpoints of $BC$ and $BD$ respectively), and from the congruence of $MAB$ and $NAB$ we have $\angle{CBA} = \angle{MBA} = \angle{NBA} = \angle{DBA}$, and again $|AB|$ is equal to itself. Thus $ABC$ and $ABD$ are congruent, thus $\mathbf{|AC| = |AD|}$ and $\mathbf{|BC| = |BD|}$.

By applying the same logic to faces $BCD$ and $ACD$ we get $|AC| = |BC|$ and $|AD| = |BD|$. Finally, applying the same logic to faces $ACB$ and $ACD$ we get $|AB| = |AD|$ and $|CB| = |CD|$. Putting all these equalities together, we get that all edges of the tetrahedron are equal, and thus the tetrahedron is regular. $\blacksquare$

See Also

1980 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_USAMO_Problems/Problem_4&oldid=191126"