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3 votes
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Find the value of the tangent segment PC in the circle below

Note that $OPCH$ and $OCTH$ are cyclic quadrilaterals. Therefore using Ptolemy's theorem on each, we can derive the equations: $$OP\cdot CH+OH\cdot PC=PH\cdot OC$$ $$13CH+OH\cdot PC=15OC\tag1$$ $$HT\...
ACB's user avatar
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2 votes
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Find the segment BT in the triangle inscribed below

Note first of all that chords $BM$ and $AB$ subtend the same angles in their respective circles, and the same holds for $BN$ and $BC$. Hence: $$ {BM\over AB}={BN\over BC}={r\over R} \quad\text{and}\...
Intelligenti pauca's user avatar
2 votes
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Calculate the segment $AC$ in the secant circles below

Since $\angle AMC=\angle ANC=90^o$ then points A, M, N and C are on a circle on diameter AC. For the same reason point B, M , K and N are on a circle on diameter BK. Power of A to circle t: $AR^2=AM\...
sirous's user avatar
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1 vote

What are the Chebyshev sets for the taxicab metric?

Surprisingly to me, it appears that $S$ need not be convex! Let $\| \cdot \|$ be any norm on $\mathbb{R}^n$ and consider the induced metric $d(x, y) = \| x - y \|$. I have the following geometric ...
Qiaochu Yuan's user avatar
1 vote

Is it true that for any real valued function $f(t)$ on $\mathbb{R},$ $f(t)a + (1-f(t))b$ is also a line contained in $L$?

In the following, we assume $A \ne B$. Denote $S = \{f (t) A + (1 - f (t)) B | t \in \mathbb{R}\}$. I claim that $S$ is a subset of the line $L = \{t A + (1 - t) B | t \in \mathbb{R}\}$ joining $A$ ...
K. Jiang's user avatar
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1 vote
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Am I incorrectly interpreting axiom I, 3 of Hilbert's Foundations of Geometry?

Claim: For every pair of points $A, B$ there exists a point $C$ not on $AB$. Proof: Suppose there does not exist any such point $C$. Then all points lie on $AB$. Then there is no set of three points ...
Numeral's user avatar
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1 vote

Find the segment BT in the right triangle below

$ \angle GHC = 90^o \\ \triangle EBC \sim \triangle GHC\\ \angle CFH =180^o -\angle HGC\\ \angle AE H +\angle HFA = 180^o \implies AEHF_{inscr.}\\ BT^2 = BH.BF = BE.BA = 36 \therefore\boxed{ BT = 6}$
peta arantes's user avatar
  • 6,645
1 vote

Find the segment BT in the right triangle below

The answer was initially given as 3, which is incorrect. This GeoGebra sketch confirms the amended answer of 6. This does not provide a solution, but may be useful nonetheless. Observe that $\overline{...
Daniel Mathias's user avatar
1 vote
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Find the segment BT in the right triangle below

Comment: As can be seen in figure BC is not necessarily equal to AB. IF BG is 3(as answer says) then you need to show that the intersection of a circle with radius 3 with circle s is point G and ...
sirous's user avatar
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