Found problems: 619
A tetrahedron $ABCD$ is given. The incircles of triangles $ ABC$ and $ABD$ with centers $O_1, O_2$, touch $AB$ at points $T_1, T_2$. The plane $\pi_{AB}$ passing through the midpoint of $T_1T_2$ is perpendicular to $O_1O_2$. The planes $\pi_{AC},\pi_{BC}, \pi_{AD}, \pi_{BD}, \pi_{CD}$ are defined similarly. Prove that these six planes have a common point.
The tetrahedron $OPQR$ has the $\angle POQ = \angle POR = \angle QOR = 90^o$. $X, Y, Z$ are the midpoints of $PQ, QR$ and $RP.$ Show that the four faces of the tetrahedron $OXYZ$ have equal area.
Suppose that Romeo and Juliet each have a regular tetrahedron to the vertices of which some positive real numbers are assigned. They associate each edge of their tetrahedra with the product of the two numbers assigned to its end points. Then they write on each face of their tetrahedra the sum of the three numbers associated to its three edges. The four numbers written on the faces of Romeo's tetrahedron turn out to coincide with the four numbers written on Juliet's tetrahedron. Does it follow that the four numbers assigned to the vertices of Romeo's tetrahedron are identical to the four numbers assigned to the vertices of Juliet's tetrahedron?
Let $ABCD$ be a tetrahedron and let $E,F,G,H,K,L$ be points lying on the edges $AB,BC,CD$ $,DA,DB,DC$ respectively, in such a way that
\[AE \cdot BE = BF \cdot CF = CG \cdot AG= DH \cdot AH=DK \cdot BK=DL \cdot CL.\]
Prove that the points $E,F,G,H,K,L$ all lie on a sphere.
Prove that if in a tetrahedron $ ABCD $ opposite edges are equal, i.e. $ AB = CD $, $ AC = BD $, $ AD = BC $, then the lines passing through the midpoints of opposite edges are mutually perpendicular and are the axes of symmetry of the tetrahedron.
In the tetrahedron $ ABCD $ from the vertex $ A $, the perpendiculars $ AB '$, $ AC' $ are drawn, $ AD '$ on planes dividing dihedral angles at edges $ CD $, $ BD $, $ BC $ in half. Prove that the plane $ (B'C'D ') $ is parallel to the plane $ (BCD) $.
Let $ABCD$ be the regular tetrahedron, and $M, N$ points in space. Prove that: $AM \cdot AN + BM \cdot BN + CM \cdot CN \geq DM \cdot DN$
Let $ABCD$ be a tetrahedron with $AB=41$, $AC=7$, $AD=18$, $BC=36$, $BD=27$, and $CD=13$, as shown in the figure. Let $d$ be the distance between the midpoints of edges $AB$ and $CD$. Find $d^{2}$.
[asy]
pair C=origin, D=(4,11), A=(8,-5), B=(16,0);
draw(A--B--C--D--B^^D--A--C);
draw(midpoint(A--B)--midpoint(C--D), dashed);
label("27", B--D, NE);
label("41", A--B, SE);
label("7", A--C, SW);
label("$d$", midpoint(A--B)--midpoint(C--D), NE);
label("18", (7,8), SW);
label("13", (3,9), SW);
pair point=(7,0);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));[/asy]
Prove that every tetrahedron has a vertex whose three edges have the right lengths to form a triangle.
The insphere of any tetrahedron has radius $r$. The four tangential planes parallel to the side faces of the tetrahedron cut from the tetrahedron four smaller tetrahedrons whose in-sphere radii are $r_1, r_2, r_3$ and $r_4$. Prove that $$r_1 + r_2 + r_3 + r_4 = 2r$$
Prove that the sum of the six angles subtended at an interior point of a tetrahedron by its six edges is greater than 540°.
A tetrahedron $ABCD$ is given such that $AD = BC = a; AC = BD = b; AB\cdot CD = c^2$. Let $f(P) = AP + BP + CP + DP$, where $P$ is an arbitrary point in space. Compute the least value of $f(P).$
It is given a tetrahedron with vertices $A,B,C,D$.
(a) Prove that there exists a vertex of the tetrahedron with the following property: the three edges of that tetrahedron through that vertex can form a triangle.
(b) On the edges $DA,DB$ and $DC$ there are given the points $M,N$ and $P$ for which:
$$DM=\frac{DA}n,\enspace DN=\frac{DB}{n+1}\enspace DP=\frac{DC}{n+2}$$where $n$ is a natural number. The plane defined by the points $M,N$ and $P$ is $\alpha_n$. Prove that all planes $\alpha_n$, $(n=1,2,3,\ldots)$ pass through a single straight line.
A paper tetrahedron is cut along some of so that it can be developed onto the plane. Could it happen that this development cannot be placed on the plane in one layer?
Let $ABCD$ is a tetrahedron. Show that
a)If $AB=CD,AC=DB,AD=BC$ then triangles $ABC,ABD,ACD,BCD$ are acute.
b)If the triangles $ABC,ABD,ACD,BCD$ have the same area , then $AB=CD,AC=DB,AD=BC$.
(I.Bogdanov, 11) Let $ h$ be the least altitude of a tetrahedron, and $ d$ the least distance between its opposite edges. For what values of $ t$ the inequality $ d>th$ is possible?
Prove that the sum of the six angles subtended at an interior point of a tetrahedron by its six edges is greater than 540°.
An equilateral triangle has side length $ 6$. What is the area of the region containing all points that are outside the triangle and not more than $ 3$ units from a point of the triangle?
$ \textbf{(A)}\ 36\plus{}24\sqrt{3} \qquad
\textbf{(B)}\ 54\plus{}9\pi \qquad
\textbf{(C)}\ 54\plus{}18\sqrt{3}\plus{}6\pi \qquad
\textbf{(D)}\ \left(2\sqrt{3}\plus{}3\right)^2\pi \\
\textbf{(E)}\ 9\left(\sqrt{3}\plus{}1\right)^2\pi$
The circumcenter and incenter of a given tetrahedron coincide. Prove that all its faces are congruent.
The tetrahedron $ S.ABC$ has the faces $ SBC$ and $ ABC$ perpendicular. The three angles at $ S$ are all $ 60^{\circ}$ and $ SB \equal{} SC \equal{} 1$. Find the volume of the tetrahedron.
Is it true that of the four heights of an arbitrary tetrahedron, three can be selected from which a triangle can be made?
Prove that if $h_1,h_2,h_3,h_4$ are the altitudes of a tetrahedron and $d_1,d_2,d_3$ the distances between the pairs of opposite edges of the tetrahedron, then
$$\frac{1}{h_1^2}
+\frac{1}{h_2^2}
+\frac{1}{h_3^2}
+\frac{1}{h_4^2}
=\frac{1}{d_1^2}
+\frac{1}{d_2^2}
+\frac{1}{d_3^2}.$$
Through a random point $C_1$ from the edge $DC$ of the regular tetrahedron $ABCD$ is drawn a plane, parallel to the plane $ABC$. The plane constructed intersects the edges $DA$ and $DB$ at the points $A_1,B_1$ respectively. Let the point $H$ is the midpoint of the altitude through the vertex $D$ of the tetrahedron $DA_1B_1C_1$ and $M$ is the center of gravity (barycenter) of the triangle $ABC_1$. Prove that the measure of the angle $HMC$ doesn’t depend on the position of the point $C_1$. [i](Ivan Tonov)[/i]
Points $P,Q,R,S$ are taken on respective edges $AC$, $AB$, $BD$, and $CD$ of a tetrahedron $ABCD$ so that $PR$ and $QS$ intersect at point $N$ and $PS$ and $QR$ intersect at point $M$. The line $MN$ meets the plane $ABC$ at point $L$. Prove that the lines $AL$, $BP$, and $CQ$ are concurrent.
Prove that the sum of the face angles at each vertex of a tetrahedron is a straight angle if and only if the faces are congruent triangles.